GRAPHIC    ALGEBRA 


s 


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^^y&fe- 


THE  MACMILLAN  COMPANY 

NEW  YORK    ■    BOSTON   •    CHICAGO 
ATLANTA   •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON  •    BOMBAY   •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


GRAPHIC    ALGEBRA 


BY 


ARTHUR   SCHULTZE,   Ph.D. 

ASSISTANT    PROFESSOR    OF    MATHEMATICS,    NEW    YORK    UNIVERSITY 

HEAD    OF    THE    DEPARTMENT    OF    MATHEMATICS,    HIGH 

SCHOOL    OF    COMMERCE,    NEW    YORK 


THE   MACMILLAN   GOMPANY 

1909 


All  rights  reserved 


S3 


Copyright,  190S, 
By  THE  MACMILLAN  COMPANY. 


Set  up  and  electrotyped.     Published  February,  1908.     Reprinted 
January,  1909. 


•  •  ■ 

...         •     • 


»   •   * 


■    * 


•  •      • 


•   •  •  .•  •  •   • 


•  •  . 


•  -  * 


>  *  •  • , 

*•. • 


NorfoooD  $wsa 

J.  8.  Cushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

It  is  now  generally  conceded  that  graphic  methods  are  not 
only  of  great  importance  for  practical  work  and  scientific 
investigation,  but  also  that  their  educational  value  for  sec- 
ondary instruction  is  very  considerable.  Consequently,  an 
increasing  number  of  schools  have  introduced  graphic  algebra 
into  their  courses,  and  several  elementary  text-books  on  graphs 
have  been  published. 

This  book  gives  an  elementary  presentation  of  all  the  funda- 
mental principles  included  in  such  courses,  and  contains  in 
addition  a  number  of  methods  which  are  shorter  and  require 
less  numerical  work  than  those  usually  given.  Thus,  for  the 
solution  of  a  cubic  or  biquadratic  by  the  customary  method 
a  great  deal  of  calculation  is  necessary  to  determine  the  co- 
ordinates of  a  number  of  points.  To  avoid  these  calculations 
and  to  make  the  work  truly  graphic,  the  author  has  devised  a 
series  of  methods  for  solving  quadratics,  cubics,  and  biquad- 
ratics by  means  of  a  standard  curve  and  straight  lines  or 
circles. 

Two  of  these  methods  —  the  solution  of  quadratics  by  a 
parabola  (§  30)  and  of  incomplete  cubics  by  a  cubic  parabola 
(§  49)  —  are  but  slight  modifications  of  methods  previously 
known ;  the  others  are  original  with  the  author,  who  first  pub- 
lished them  in  a  paper  read  before  the  American  Mathematical 
Society  in  April,  1905. 

The  author  desires  to  acknowledge  his  indebtedness  to 
Mr.  Charles  E.  Deppermann  for  the  careful  reading  of  the 
proofs  and  for  verifying  the  results  of  the  examples. 

ARTHUR   SCHULTZE. 
New  York,  December,  1907. 


260013 


CONTENTS 

PAET  I 
General  Graphic  Methods 


CHAPTER   I 

PAGE 

Definitions 1 


CHAPTER   II 
Graphic  Representation  of  a  Function  of  One  Variable       .        4 

CHAPTER  III 

Graphic    Solution    of    Equations    involving    One    Unknown 

Quantity    ...........       13 

CHAPTER  IV 

Graphic    Solution    of    Equations    involving    Two    Unknown 

Quantities 17 


PAET   II 

Solution  of  Equations  by  Means  of  Standard 

Curves 

CHAPTER  V 

Quadratic  Equations 26 

vii 


Vlll  CONTENTS 

CHAPTER  VI 

PAGE 

Cubic  Equations 42 

CHAPTER  VH 
Biquadratic  Equations 69 

APPENDIX 

I.     Graphic  Solution  of  Problems 73 

II.     Statistical  Data  Suitable  for  Graphic  Representation  78 

III.     Tables 84 

Answers  to  Exercises 89 


GRAPHIC    ALGEBRA 


it       i   > 


PAET  I 


GENERAL  GRAPHIC  METHODS 


CHAPTER  I 


DEFINITIONS 


A 


N 


A" 


1.  Location  of  a  point.  If  two  fixed  straight  lines  XX'  and 
YY  meet  in  0  at  right  angles,  and  PM  _L  XX',  and  PN±  YY, 
then  the  position  of  the  point  P  is  determined  if  the  lengths 
of  PM  and  PN  are  given. 

2.  Coordinates.  The 
lines  PN  and  PM  are 
called  the  coordinates  of 
point  P;  PN,  or  its  equal 
OM,  is  the  abscissa;  and 
PM,  or  its  equal  ON,  is 
the  ordinate  of  point  P. 
The  abscissa  is  usually- 
denoted  by  x,  the  ordinate 

by  y- 

The  line  XX'  is  called 
the  jr-axis  or  the  axis  of 
abscissas,  YY'  the  y-zxis 
or  the  axis  of  ordinates. 
The  point  0  is  the  origin,  and  M  and  N  are  the  projections  of 
P  upon  the  axes.  Abscissas  measured  to  the  right  of  the  ori- 
gin, and  ordinates  above  the  x-axis,  are  considered  positive; 
hence  coordinates  lying  in  opposite  directions  are  negative. 


-(P 


&         3  M4 


D 


2 


GRAPHIC  ALGEBRA 


3.  The  point  whose  abscissa  is  x,  and  whose  ordinate  is  y,  is 
usually  denoted  by  (x,  y).    Thus  the  points  A,  B,  C,  and  D  are 

respectively  represented 
by  (3,4),  (-2,  3),  (-3, 
-  2),  and  (2,  -  3). 

The  process  of  locating 
a  point  whose  coordinates 
are  given  is  called  plotting 
the  point. 

4.  Since  there  are  other 
methods  of  determining 
the  location  of  a  point, 
the  coordinates  used  here 
are  sometimes,  for  the 
sake  of  distinction,  called 
rectangular  coordinates. 

Note  1.  While  usually  the  same  length  is  taken  to  represent  the  unit 
of  the  ahscissas  and  the  unit  of  the  ordinates,  it  is  sometimes  convenient 
to  draw  the  x  and  the  y  on  different  scales. 

Note  2.  Graphical  constructions  are  greatly  facilitated  by  the  use  of 
cross-section  paper,  i.e.  paper  ruled  with  two  sets  of  equidistant  and  par- 
allel lines  intersecting  at  right  angles.     (See  diagram  on  page  29.) 


, 

\Y 

A 

! 

! 

! 
i 
t 

__^T. 

p 

3 
2 

1 

■u 

X' 

i 

X 

x 

'-[3 

-2 

"T" l 

) 

1          2 

3   J/4' 

i 
i 

i 
1 

-1 

C 

-& 

=3- 

i 
D 

EXERCISE  1 

1.  Plot  the  points:  (3,  2),  (4,  -1),  (-3,  2),  (-3,  -3). 

2.  Plot  the  points:  (-2,  3),  (-5,  0),  (4,  -3),  (0,  4). 

3.  Plot  the  points  :  (0,  -  4),  (4,  0),  (0,  0). 

4.  Draw  the  triangle  whose  vertices  are  respectively  (4, 1), 
(-1,3),  and  (1,  -2). 

5.  Plot  the  points  (—2, 1)  and  (2,  —3),  and  measure  their 
distance. 

6.  What  is  the  distance  of  the  point  (3,  4)  from  the  origin? 

7.  Where  do  all  points  lie  whose  ordinates  equal  4? 


DEFINITIONS  3 

8.  Wliere  do  all  points  lie  whose  abscissas  equal  zero  ? 

9.  "Where  do  all  points  lie  whose  ordinate  equals  zero? 

10.  What  is  the  locus  of  (x,  y)  if  y  =  3? 

11.  If  a  point  lies  in  the  o^axis,  which  of  its  coordinates  is 
known? 

12.  What  are  the  coordinates  of  the  origin? 


CHAPTER  II 

GRAPHIC  REPRESENTATION  OF  A  FUNCTION  OF  ONE 

VARIABLE 

5.  Definitions.  An  expression  involving  one  or  several  let- 
ters is  called  a  function  of  these  letters. 

x2  —  x  +  7  is  a  function  of  x. 

q 

Vy y2  is  a  function  of  y. 

y 

2  xy  —  y2  +  3  y3  is  a  function  of  x  and  y. 

If  the  value  of  a  quantity  changes,  the  value  of  a  function 
of  this  quantity  will  change,  e.g.  if  x  assumes  successively  the 
values  1,  2,  3,  4,  x2  —  x  +  1  will  respectively  assume  the  values 
7,  9,  13,  19.  If  x  increases  gradually  from  1  to  2,  x2  —  x  +  7 
will  change  gradually  from  7  to  9. 

A  variable  is  a  quantity  whose  value  changes  in  the  same 
discussion. 

A  constant  is  a  quantity  whose  value  does  not  change  in  the 
same  discussion. 

In  the  example  of  the  preceding  article,  x  is  supposed  to  change,  hence 
it  is  a  variable,  while  7  is  a  constant. 

6.  Temperature  graph.  A  convenient  method  for  the  repre- 
sentation of  the  various  values  of  a  function  of  a  letter,  when 
this  letter  changes,  is  the  method  of  representing  these  values 
graphically ;  that  is,  by  a  diagram.  This  method  is  frequently 
used  to  represent  in  a  concise  manner  a  great  many  data  refer- 
ring to  facts  taken  from  physics,  chemistry,  technology,  eco- 
nomics, etc. 

To  give  first  an  example  of  one  of  these  applications,  let 
us  suppose   that  we  have   measured  the  temperatures  at  all 

4 


FUNCTION  OF  ONE   VARIABLE 


hours,  from  12  m.  to  11  p.m.,  on  a  certain  day,  and  that  we 
have  found : 


At  12  m. 

3°C. 

At  6  p.m. 

5°  a 

At  1  P.M. 

5°C. 

At  7  p.m. 

3£°C. 

At  2  p.m. 

6i°  C. 

At  8  p.m. 

2°  a 

At  3  p.m. 

7°  C. 

At  9  p.m. 

2     v* 

At  4  p.m. 

6f°C. 

At  10  p.m. 

-1°C. 

At  5  p.m. 

6°  a 

At  11  P.M. 

-2i°C 

>• 

7° 

b 

fl° 

V 

i 

( 

1 

r° 

n 

1° 

17 

[  - 

rt° 

4 

1 

0° 

1° 

i 

X 

0 

> 

)      ; 

i     i 

I     ; 

)    f 

r  i 

> 

)  10  1 

1  P 

M. 

i 

..V 

To  represent  graphically  one  of  these  facts,  e.g.  the  tem- 
perature at  6  p.m.  was  5°,  construct  a  point  G,  whose  abscissa 
is  6  and  whose  ordinate  is 
5,  taking  any  convenient 
lengths  as  uuits.  Eepre- 
senting  in  a  similar  way 
the  temperatures  at  all 
hours,  we  obtain  the  points 
(0,  3),  (1,  5),  (2,  61),  (3,  7), 
etc.,  i.e.  A,  B,  G,  D,  .  .  .  M. 

The  diagram  thus  con- 
structed contains  all  the 
information  given  in  the  table,  but  it  gives  it  in  a  clear  and 
concise  form,  that  at  once  impresses  upon  the  eye  the  relative 
values  of  the  temperatures  and  their  changes. 

In  a  similar  manner  we  may  plot  the  temperatures  at  any 
time  between  12  m.  and  11  p.m.  Thus,  to  represent  the  fact 
that  the  temperature  at  1.30  p.m.  was  6°,  construct  the  point 
(Xh  6)- 

7.  If  we  represented  the  temperatures  of  every  moment 
between  12  m.  and  11  p.m.,  we  would  obtain  an  uninterrupted 
sequence  of  points,  or  a  curved  line,  as  shown  in  the  next  dia- 
gram. This  curve  is  said  to  be  a  graphical  representation  or  a 
graph  of  the  temperatures  from  12  m.  to  11  p.m.  It  is,  of 
course,  not  possible  to  construct  an  infinite  number  of  points, 


GRAPHIC  ALGEBRA 


hence  every  graph  constructed  in  the  above  manner  is  only  an 
approximation,  whose  accuracy  depends  upon  the  number  of 
points  constructed. 

To  find  from  the  diagram  the  temperature  at  any  time,  e.g. 
at  2.30,  measure  the  ordinate  which  corresponds  to  the 
abscissa  21;  to  find  when  the  temperature  was  4°,  measure 
the  abscissa  that  corresponds  to  the  ordinate  4,  etc. 


Y 

0 

a 

1 

1 

1 

X 

0 

i 

2 

1    • 

\ 

3     1 

' 

i 

y\ 

0  1 

1  p.  d. 

EXERCISE  2 

1.    From  the  diagram  find  approximate  answers  to  the  follow- 
ing questions : 

a.  Determine  the  temper- 
ature at : 

5  p.m.,  1.30  p.m.,  5.45  p.m., 
11.45  a.m. 

b.  At  what  hour  or  hours 
was  the  temperature  6°,  5°, 
1°,  -  1°,  0°  ? 

c.  At  what  hour  was  the 
temperature  highest  ? 

d.  What  was  the  highest  temperature  ? 

e.  During  what  hours  was  the  temperature  above  5°  ? 

/.  During  what  hours  was  the  temperature  between  3°  and 
4°? 

g.  During  what  hours  was  the  temperature  above  0°  ? 

h.   During  what  hours  was  the  temperature  below  0°  ? 

i.   How  much  higher  was  the  temperature  at  4  than  at  8  p.m.  ? 

k.   At  what  hour  was  the  temperature  the  same  as  at  1  p.m.  ? 

I.  During  what  hours  did  the  temperature  increase  ? 

m.   During  what  hours  did  the  temperature  decrease  ? 

n.  Between  which  two  successive  hours  did  the  temperature 
change  least  ? 

o.  Between  which  two  successive  hours  did  the  temperature 
increase  most  rapidly  ? 


FUNCTION  OF  ONE   VARIABLE  7 

2.    Construct  a  diagram  containing  the  graphs  of  the  mean 
temperatures  of  the  following  four  cities : 


^H 

i-t 

T-H 

rH 

rH 

T— 1 

i-H 

th 

H 

i—i 

tH 

a 

•4 

£ 

a 
w 

PS 

< 

3 

1* 

< 

P 

o 
p 

O 

5 

d 

a 

1-5 

fc* 

S 

«f 

y 

l-B 

i-s 

<f 

CD 

O 

to 
43 

R 
34 

H 

New  York  City 

30 

32 

37 

48 

60 

69 

74 

72 

66 

55 

52 

San  Francisco 

50 

52 

54 

55 

57 

58 

58 

59 

60 

59 

56 

51 

56 

Tampa 

59 

66 

66 

72 

76 

80 

82 

81 

80 

73 

65 

63 

72 

Bismarck 

4 

10 

23 

42 

54 

64 

70 

68 

57 

44 

26 

15 

40 

a.  Which  of  these  cities  has  the  most  uniform  temperature  ? 

b.  Which  one  has  the  greatest  extremes  of  temperature  ? 

c.  When  is  the  mean  temperature  in  San  Francisco  the 
same  as  in  New  York  ? 

d.  When  does  the  mean  temperature  of  New  York  rise  most 
rapidly  ? 

e.  What  is  the  difference  between  the  mean  temperatures  of 
New  York  and  Bismarck  on  Jan.  15  ? 

3.  By  using  the  annexed  table  represent  graphically  the 
greatest  amount  of  water  vapor  which  a  cubic  meter  of  air  can 
hold  at  various  temperatures. 


Degrees  of  Centigrade 

—  25 

—  20 

—  15 

-10 

—  5 

0 

5 

10 

15 

20 

25 

3n 
30 

35 
39.3 

40 

Grams  of  Vapor 

.8 

1.0 

1.5 

2.3 

3.4 

4.8 

6.9 

:>.:: 

12.S 

17.1 

■23.0 

50.6 

a.  Represent  graphically  by  a  point  air  of  30°  C.  which  holds 
15  grams  of  water  vapor  per  cubic  meter. 

b.  If  such  air  would  cool,  represent  the  change  graphically 
by  a  line. 

c.  At  what  temperature  would  such  air  become  saturated, 
i.e.  contain  all  the  moisture  it  can  hold?* 


*  This  temperature  is  called  the  dew-point. 


8 


GRAPHIC  ALGEBRA 


d.  If  the  same  air  was  cooled  to  5°,  how  many  grams  of 
moisture  would  be  condensed  per  cubic  meter  ? 

e.  How  much  more    moisture  per   cubic  meter  can  air  of 
the  kind  mentioned  in  Ex.  a  hold  ?  * 

[For  more  statistical  data  suitable  for  graphic  representation 
see  Appendix  II.] 


8.  Graph  of  a  function.  The  values  of  a  function  for  the  va- 
rious values  of  x  may  be  given  in  the  form  of  a  numerical  table. 
Thus  the  table  on  page  84  gives  the  values  of  the  functions 

r-  1 
x2,  x3,  V as,  -  for  x  =  1,  2,  3,  •  •  •  up  to  100.      The  values  of 

functions  may,  however,  be  also  represented  by  a  graph. 
E.g.  to  construct  the  graph  of  x2  construct  a  series  of  points 

whose  abscissas  represent 
x,  and  whose  ordinates  are 
x%  i.e.  construct  the  point 
(-3,9),  (-2,4),  (-1,1), 
(0,  0)  •  •  •  (3,  9),  and  join 
the  points  in  order. 

If  a  more  exact  diagram 
is  required,  plot  points 
which  lie  between  those 
drawn  above,  as  (^,  \), 
C4,  21),  etc. 

Since  the  squares  of  the  numbers  increase  very  rapidly,  it  is  convenient 
to  make  the  scale  unit  of  the  x2  smaller  than  that  of  the  x.  The  graph 
on  page  29  was  constructed  in  this  manner. 

To  find  from  the  graph  the  square  of  —  2.5,  measure  the 
ordinate    corresponding   to  the  abscissa  —2.5,  i.e.  6.25.     To 

*  Many  meteorological  facts  can  be  explained  by  the  graph  of  Ex.  3, 
e.g.  the  meaning  of  "dew-point,"  relative  and  absolute  humidity,  the  fact 
that  the  mixing  of  two  masses  of  saturated  air  of  different  temperatures 
produces  precipitation,  etc. 


L_ 

i\ 

p- 

\ 

/ 

\ 

i 

/ 

\ 

/ 

\ 

/ 

\ 

\ 

\ 

s 

V 

y 

X 

f  -2  +  fl 

L      2 
T 

<d 

FUNCTION  OF  ONE   VARIABLE 


find  VT,  measure  the  abscissa  whose  ordinate  is  7,  i.e.  +2.6 
or  -  2.6. 

Ex.  Draw  the  graph  of  i  x2  —  A  x  —  3. 

To  obtain  the  values  of  the  functions  for  the  various  values 
of  x,  the  following  arrangement  may  be  found  convenient : 

(Compute  each  column  before  commencing  the  next,  and  see  table  on 
page  84.) 


X 

X2 

t*2 

-i* 

■^x-  —  -g  X  —  O 

-  4 

16 

8 

.8 

5.8 

-3 

9 

4.5 

.6 

2.1 

-2 

4 

2 

.4 

-   .6 

-  1 

1 

.5 

.2 

-2.3 

0 

0 

0 

0 

-3 

1 

1 

.5 

-.2 

-2.7 

2 

4 

2 

-.4 

-1.4 

3 

9 

4.-3 

-.6 

.9 

4 

16 

8 

-.8 

4.2 

Locating  the   points  (-4,5.8),   (-3,2.1),   (- 2,  -  .6),  ••-,   (4,4.2), 
and  joining  in  order  produces  the  graph  ABC. 

9.    For  brevity,  the  function  is  frequently  represented  by  a 
single  letter,  as  y.     Thus,  in  the  above 
example, 


i 


y  =  ix~-i 


x- 


3; 


if  #  =  i,  we  fiud  from  the  graph 
1  x2  —  i  x  —  3  or  y  =  —  3,  if  x  —  1\, 
y  =  —  .4,  etc. 


For  values  of  x  greater  than  4,  the  func- 
tion will  obviously  be  always  positive  and 
increase  when  x  increases.  Hence  the  curve 
will  continue  to  go  upward  beyond  C,  and 
similarly  above  A. 

Graphs  should  always  be  drawn  until  they  reach  their  ultimate  direc- 
tion at  both  ends. 


\A 

Y' 

, 

\ 

1 

\ 

\ 

\ 

X' 

\ 

t 

X 

i  • 

3  \ 

> 

l 

. 

11 

i    i 

\ 

-1 

0 

1 

V 

r: 

B 

^4- 

10  GRAPHIC  ALGEBRA 

10.    The  graph  of  an  equation  of  the  form  of  ax?  +  bx  +  c  is 
called  a  parabola. 

Thus  the  graph  of  \  x2  —  \  x  —  3  is  a  parabola. 


EXERCISE  3 

Draw  the  gr; 

aphs 

of  the  following  functions  : 

# 

1.  x  +  2. 

9.  x2-l. 

17. 

X2  —  X  +  1. 

2.  3  z  +  5. 

10.    X2  +  05. 

18. 

6  +05  — aA 

3.  2  as -7. 

11.  x2-2x. 

19. 

2-as-aj2. 

4.  fa;. 

12.  4  a;  — ar8. 

20. 

10  -  3  x  -  x2. 

5.  1  — re. 

13.  a?2  —  4*4-4. 

21. 

2  x2  +  5  x  -  20 

6.  2-3a,\ 

14.  x2  —  x  —  5. 

22. 

ar3. 

7.   —  3  as. 

15.  as* -3  a; -8. 

23. 

a,*3  —  2  a5. 

8.   \tf. 

16.  a?*  +  a;  — 2. 

24. 

ar3  —  a;  +  1. 

25.  Draw  the  graph  of  x2  from  05  =  —  4  to  a;  =  4,  and  from 
the  diagram  find : 

a.  (3.5)2;     6.  (-1.6)*;     c.  (-2.8)2;     d.  (1.9)2; 

e.    V6\25;     /.   Vl2^25;     gr.    V5 ;     ft.   V^3. 

26.  Draw  the  graph  of  x2  —  4  05  +  2  from  05  =  —  1  to  a;  =  4, 
and  from  the  diagram  determine : 

(a)  The  values  of  the  function  if  as  =  —  ^,  1^,  2^. 
(6)  The  values  of  as,  if  x2  —  4  a;  +  2  equals  —  2,  1,  1£. 

(c)  The  smallest  value  of  the  function. 

(d)  The  value  of  x  that  produces  the  smallest  value  of  the 
function : 

(e)  The  values  of  a;  that  make  x2  —  4  x  +  2  =  0. 
(/)  The  roots  of  the  equation  05*  —  4  x  +  2  =  0. 
(p*)  The  roots  of  the  equation  as2  —  4a?  +  2=—  1. 
(ft)  The  roots  of  the  equation  ar  —  4  as  +  2  =  2. 

*  If  necessary,  use  for  the  ordinates  a  smaller  auit  than  for  the  ab- 
scissas. 


FUNCTION  OF  ONE   VARIABLE  11 

27.  Draw  the  graph  oiy  =  2 -\-2x-x2,  from  x  =  —  2  to  x  =  4, 
and  from  the  diagram  determine  : 

(a)  The  values  of  y,  i.e.  the  function,  if  x—%,  —  1£,  2\. 

(b)  The  values  of  x  if  y  =  —  2. 

(c)  The  greatest  value  of  the  function. 

(d)  The  value  of  a;  that  produces  the  greatest  value  of  y. 

(e)  The  values  of  x  if  the  function  equals  zero. 
(/)  The  roots  of  the  equation  2  +  2  x  -  x2  =  0. 
(#)  The  values  of  x  if  y  =  1. 

(/i)  The  roots  of  the  equation  2  +  2cc— a^  =  l. 

28.  The  formula  for  the  distance  traveled  by  a  falling  body- 
is  S  =  ±gt2. 

(a)  Eepresent  i  gt2  graphically  from  t  =  0  to  t  =  5.  (Assume 
g  =  10  meters,  and  make  the  scale  unit  of  the  t  equal  to  10 
times  the  scale  unit  of  the  ^  gt2.) 

(b)  How  far  does  a  body  fall  in  2\  seconds  ? 

(c)  In  how  many  seconds  does  a  body  fall  25  meters  ? 


11.  A  function  of  the  first  degree  is  an  integral  rational  func- 
tion involving  only  the  first  power  of  the  variable. 

Thus,  4x  +  7orax  +  6  +  c  are  functions  of  the  first  degree. 

12.  It  can  be  proved  that  the  graph  of  a  function  of  the  first 
degree  is  a  straight  line,  hence  two  points  are  sufficient  for  the 
construction  of  these  graphs.  (This  is  true  if  the  abscissas  and 
ordinates  are  drawn  on  different  scales  or  on  the  same  scale.) 

It  can  easily  he  shown  that  the  preceding  proposition  is  true  for  any 
particular  example,  e.g.  3  x  +  2. 

If  a;  =-3,  -2,  -1,  0,  1,  2,  3, 

then  3x  +  2=-7,  —  4,  —  1,  2,  5,  8,  11; 

i.e.  if  x  increases  hy  1,  3  x  +  2  increases  hy  3.  Hence  if  a  straight  line 
be  drawn  through  (—3,-7)  and  (  —  2,  —  4),  this  line  will  ascend  3  units 
from  x  =  —  3  to  x  =  —  2.  Obviously  the  prolongation  of  this  line  will 
ascend  at  the  same  rate  throughout,  and  it  will  pass  through  (—  1,  —  1), 
(0,  2),  etc. 


12  GRAPHIC  ALGEBRA 

Instead  of  plotting  (—3,  —7)  and  (—2,  —4),  any  other  two  points 
may  be  taken.  It  is  advisable  not  to  select  two  points  which  lie  very 
closely  together. 

EXERCISE   4 
Draw  the  graph  of 

1.   3  a? -10.  3.   2  a;— 7.  5.   6+ a. 

2.5^  +  2.  4.  2  —  3  x.  6.  |  x  —  5. 

7.  Degrees  of  the  Fahrenheit  scale  are  expressed  in  degrees 
of  the  Centigrade  scale  by  the  formula  C.  =  -|  (F.  —  32). 

(a)  Draw  the  graph  of  f  (F.  -  32),  from  F.  =  -  5,  to  F.  =  40. 
(6)  From  the  diagram  find  the  number  of  degrees  of  Centi- 
grade equal  to  - 1°  F.,  9°  F.,  14°  F.,  32°  F. 

(c)  Change  to  Fahrenheit  readings:  -  10° C,  0°  C,  1°  C. 

8.  Show  that  the  graphs  of  3  x  +  2  and  3  x  —  1  are  parallel 
lines. 


CHAPTER  III 


GRAPHIC    SOLUTION    OF    EQUATIONS    INVOLVING    ONE 
UNKNOWN  QUANTITY 

13.  Degree  of  an  equation.  A  rational  integral  equation  which 
contains  the  nth.  power  of  the  unknown  quantity,  but  no 
higher  power,  is  called  an  equation  of  the  7ith  degree. 

x5  —  5  x3  +  2  x2  —  7  =  0  is  an  equation  of  the  fifth  degree. 
Xs  —  2  x2  —  5  x  +  1  =  0  is  an  equation  of  the  third  degree. 

A  quadratic  equation  is  an  equation  of  the  second  degree. 

A  cubic  equation  is  an  equation  of  the  third  degree. 

A  biquadratic  equation  is  an  equation  of  the  fourth  degree. 

x3  +  2x  +  3=0isa  cubic  equation. 

x4  +  3  x3  +  2  x  —  7  =  0  is  a  biquadratic  equation. 

14.  Solution  of  equations.  Since  we  can  graphically  deter- 
mine the  values  of  x  that  make  a  function  of  x  equal  to  zero, 
it  is  evidently  possible  to  find  graphically  the  real  roots  of 
an  equation. 

Ex.    Find  graphically  the  real  roots  of  the  equation 

!B84-aj8-9a!-7  =  0. 

(In  computing  the  values  of  y  use  table  on  page  84.) 


X 

X2 

Xs 

—  9a; 

x3  +  x2  —  9x 

x3+x2— 9x—  7  or  y 

-4 

16 

-64 

36 

-12 

-19 

-3 

9 

-27 

27 

9 

2 

-2 

4 

-    8 

18 

14 

7 

-1 

1 

-    1 

9 

9 

2 

0 

0 

0 

0 

0 

-    7 

1 

1 

1 

-    9 

—  7 

-14 

2 

4 

8 

-18 

-  6 

-13 

3 

9 

27 

-27 

9 

2 

4 

16 

64 

-36 

44 

37 

13 


14 


GRAPHIC  ALGEBRA 


Obviously  the  values  of  the  f uuction  for  x  >  4  will  increase 
rapidly,  and  for  values  of  x  <  —  4  will  be  less  than  — 19. 
Locating  the  points  (-4,  -19),  (-3,  2)  (-2,  7)  ...  (4,  37) 

and  joining  produces  the 
graph  ABC. 

Since  ABC  intersects 
the  avaxis  at  three  points, 
P,  P',  and  P",  three 
values  of  x  make  the 
function  zero.  Hence 
there  are  three  roots 
which,  by  measuring  OP", 
OP',  and  OP,  are  found 
to  be  approximately  —3.1, 
-.8,  and  2.9. 

To  find  a  more  exact  answer  for  one  of  these  roots,  e.g.  OP,  we 
draw  the  portion  of  the  diagram  which  contains  P  on  a  larger  scale. 

If  x  =  2.9,  the  function  equals  —.301,  i.e.  it  is  negative.  Hence  it 
appears  from  the  diagram  that  the  roots  must  be  larger.  Substituting 
x  =  3  produces  x3  +  x'2  —  9  x  —  7  =  2,  a  positive  quantity.  The  root  there- 
fore must  lie  between  2.9  and  3. 

Making  the  unit  of  length  ten  times  as  large  as  before,  we  locate  the 
points  (2.9, —  .301)  and  (3,  2),  i.e.  B'  and  C  ,  in  diagram  II.  Since  in 
nearly  all  cases  small  portions  of  the  curve  are  almost  straight  lines, 
we  join  the  two  points  by  a  straight  line  B'C,  which  intersects  the  x-axis 
in  P. 

The  measurement  of  P  gives  the  root 

x  =  2.915. 

If  a  greater  degree  of  accuracy  is  required,  a  third  drawing  on  a  still 
larger  scale  must  be  constructed. 

15.  The  diagram  of  the  last  exercise  may  also  be  used  to 
find  the  real  roots  of  an  equation  of  the  form  a^-f  x2— 9x— 7  =  ra, 
when  m  represents  a  real  number. 

To  solve,  e.g.,  the  equation  a^  +  ar  —  9  x  —  7  =  2,  determine 
the  points  where  the  function  is  2.  If  cross-section  paper  is 
used,  the  points  may  be  found  by  inspection,  otherwise  draw 


EQUATIONS  INVOLVING  ONE  UNKNOWN  QUANTITY     15 

through  (0,  2)  a  line  parallel  to  the  £-axis,  and  determine  the 
abscissas  of  the  points  of  intersection  with  the  graph,  viz. 
-3,-1,3. 

16.  It  can  be  proved  that  every  equation  of  the  nth  degree 
has  n  roots ;  hence  if  the  number  of  the  points  of  intersection 
is  less  than  n,  the  remaining  roots  are  imaginary. 

Thus,  x*  +  x2  —  9x  —  7  =  13  has  only  one  real  root,  viz.  3.4;  hence 
two  roots  are  imaginary. 

If,  however,  the  line  parallel  to  the  x-axis  is  tangent  to 
the  curve,  the  point  of  tangency  represents  at  least  two  roots, 
and  hence  the  preceding  paragraph  cannot  be  applied. 

EXERCISE  5 

Solve  graphically  the  following  equations : 

1.  4a;  — 7  =  0.  14.   2 x2-  4 x -15=0. 

2.  2 z  +  5  =  0.  15.    2 x2  +  10 a;-7  =  0. 

3.  6-x  =  0.  16.   3ar2-6cc-13  =  0. 

4.  8-3 a;  =  0.  17.    Xs- 3^-1  =  0. 

5.  x2_a;_6  =  0  18<    ar?- 12 £  +  18  =  0. 

6.  ar-£-5  =  0.  19.    £3-4«  +  l=0. 

7.  a?  — 2 x  — 7  =  0.  20.  x3 +  £-3  =  0. 

8.  £2-6£  +  9  =  0.  21.   £3  +  3£-ll  =  0. 

9.  x2  +  5aj_4  =  o.  22.   2£3-6£  +  3=0. 

10.  £2-5£-3  =  0.  23.   £3-5a;2-9£  +  50  =  0. 

11.  x2-Sx-6  =  0.  24.   x3- 13  £-  +  38  £  +  17=0. 

12.  £2-2£-9  =  0.  25.     X4-10x2+8  =  0. 

13.  3£2-3£-17  =  0.  26.    £4-4£2  +  4x-4  =  0. 

27.  x4-6x3  +  7x2  +  6x-7  =  0. 

28.  x5  -  x4  -  11  x5  +  9  x2  + 18  x  -  4  =  0. 

29.  2x  +  x  —  4  =  0. 


16  GRAPHIC  ALGEBRA 

30.  If  y  =  xs  +  5  x2  - 10, 

(a)    Solve  y  =  0.  (c)    Solve  y  =  —  5. 

(6)   Solve  y  =  5.  (d)   Solve?/ =  -15. 

(e)   Determine  the  number  of  real  roots  of  the  equation 

y=-2. 

(/)  Determine  the  limits  between  which  m  must  lie,  iiy  =  m 
has  three  real  roots. 

(g)  Find  the  value  of  m  that  will  make  two  roots  equal  if 
y  =  m. 

(h)  Find  the  greatest  value  which  y  may  assume  for  a 
negative  x. 

(i)  Which  negative  value  of  x  produces  the  greatest  value 
ofy? 

31.  Ify=za?  —  7x  +  3, 

(a)   Solve  y  =  0.  (d)    Solve  y  =  10. 

(6)   Solve  y  =  3.  .  (e)   Solve  y  =  -15. 

(c)    Solve  y  =  —  3. 

(/)  Determine  the  number  of  real  roots  if  ?/  equals  15, 
10,  5,  or  -  7. 

(#)  Determine  the  number  of  imaginary  roots  if  y  =  — 10, 
if  ?/  =12,  if  y  =  2. 


CHAPTER   IV 


GRAPHIC    SOLUTION    OF    EQUATIONS    INVOLVING    TWO 
UNKNOWN   QUANTITIES 

17.    Graphs  of  functions  of  two  unknown  quantities.     In  §  8 

the   graph  of  the  function  \  x2  —  \x  —  3   was   discussed.     If 
£  ajS  _  £  x  —  3   is  denoted   by  y,  then 
the   ordinate   represents   the  various 
values  of  y,  and  the  annexed  diagram 
represents  the  equation 

y  =  hx*-\x-3.  (1) 

The  coordinates  of  every  point  of  the 
curve  satisfy  equation  (1),  and  every 
set  of  real  values  of  x  and  y  satisfying 
the  equation  (1)  is  represented  by 
the  coordinates  of  a  point  in  the 
curve. 

Similarly,  to  represent  — ^—  =  2  graphically  solve  for  y,  i.e. 

y-5 

x2  +  x  + 10 


\ 

y> 

v 

V 

l 

\ 

\ 

\ 

X' 

\ 

1 

X 

4    - 

3  \ 

2    - 

i  c 

. 

ll 

i   i 

\ 

-I 

K 

r; 

B 

y  =  - 


and  construct  the  graph  of 


■  • 

2 


18.  The  curve  representing  an  equation  is  called  the  graph 
or  locus  of  the  equation. 

19.  If  an  equation  containing  two  unknown  quantities  can 
be  reduced  to  the  form  y  =  f(x),  when  f(x)  represents  a  func- 
tion of  x,  then  the  equation  can  be  represented  graphically. 

17 


18 


GRAPHIC  ALGEBRA 


Ex.  1.     Eepresent   graphically  3  x  —  2  y  =  2. 

3z-2 


Solving  for  y, 


y  = 


■2 


v 

3 

.' 

<V/ 

9 

1 

0/ 

X'- 

2      - 

1 

/ 

: 

i     ; 

i   X 

/ 

/ 

r 

/ 

Y' 

Hence,   if  »  equals    —  2,  —  1,        0,  1,  2,  3  ; 
then  y  equals  —  4,  —  2£,    —  1,^,2,3$. 

Locating  the  points  (—  2,  —  4),(— 1,—  2J), 
etc.,  and  drawing  a  line  through  them,  we  ob- 
tain the  graph  of  the  equation,  which  is  a 
straight  line. 

20.  The  graph  of  an  equation  of  the 
first  degree  involving  two  unknown  quan- 
tities is  always  a  straight  v 
line,  and  hence  it  can  be  AY 
constructed  if  tioo  points 
are  located  (§  12). 
Ex.  2.     Draw  the  locus  of  4  x  +  3  y  =  12. 

Hai  =  0,y  =  4;  if  y  -  0,  x  =  3. 

Hence,  locate  points  (0,  4)  and  (3,  0),  and  join  them 
by  a  straight  line  AB.     AB  is  the  required  graph. 

Note.     Equations  of  the  first  degree  are  called  linear 
equations,  because  their  graphs  are  straight  lines. 

21.   If  two  linear  equations  differ  only  in  their 

absolute  terms  (i.e  terms  not  containing  x  or  y) 

as  2  x  +  y  =  4  and  2  x  +  y  =  2,  their  graphs  are 

parallel  lines. 

EXERCISE  6 

Draw  the  loci  of  the  following  equations  : 

1.  x  +  y  =  4:.  9.  12  x  + 15  y  =  48. 

2.  x  —  2?/ =  4.  10.  x2—  \/  +  2  =  0. 

3.  2a?  —  3y  =  12.  ll.  2x2-y- x  =  0. 

4.  a  —  ?/  =  0.  12.   a3  +  ?/  =  0. 

5.  x+y=—  10.  13.   y-  —  x  =  2. 


6.  2/  =  -  4. 

7.  .r  +  2/  =  0. 

8.  y  =  2x. 


y—x+2+l 

X  X 


14 

15.  a2    -2/2=16 


0. 


SOLUTION  OF  SIMULTANEOUS  EQUATIONS 


19 


16.  A  body  moving  with  a  uniform  velocity  of  3  yds.  per 
second  moves  in  t  seconds  a  distance  cl  =  3 1. 

Kepresent  this  formula  graphically. 

17.  If  two  variables  x  and  y  are  directly  proportional,  then 

y  =  ex,  where  c  is  a  constant. 
Show  that  the  graph  of  two  variables  that  are  directly  pro- 
portional is  a  straight  line  passing  through  the  origin  (assume 
for  c  any  convenient  number). 

18.  If  two  variables  x  and  y  are  inversely  proportional,  then 

y  —  -,  where  c  is  a  constant. 
x 

Draw  the  locus  of  this  equation  if  c  =  12. 

19.  The  temperature  remaining  the  same,  the  volume  v  of  a 
gas  is  inversely  proportional  to  the  pressure  p.  For  a  certain 
body  of  gas,  v  =  2  cubic  feet,  if  p  =  15  lbs.  per  square  inch. 
Represent  the  changes  of  p  and  v  graphically. 


22.   Graphical  solution  of  a  linear  system 
To  find  the  roots  of 
the  system : 

2x  +  3y  =  8,      (1) 
x-2y  =  2.     (2) 

By  the  method  of 
the  preceding  article 
construct  the  graphs 
AB  and  CD  of  (1) 
and  (2)  respectively. 
The  coordinates  of 
every  point  in  AB 
satisfy  the  equation 
(1),  but  only  one  point 
in  AB  also  satisfies 
equation  (2),  viz.  P, 
the  point  of  intersection  of  AB  and  CD. 

\ 


__J_  — 

,rv 

A    *v 

-ft-  —  s 

^                    '-' 

*S                                ^r 

SS,^-E>        .*2 

-                          S^ 

-^ 

^>     %       1r 

U 

__„2I         _     ^     £- 

,<"2               3              <^s 

~2                     ,*' 

^ 

3 

—1  ^^ 

^sf 

-<" 

t^                     : 

ZfZ 

L. 

_  _x 

20 


GRAPHIC  ALGEBRA 


By  measuring  the  coordinate  of  P,  we  obtain  the  roots, 
x  =  3.15,  y  =  .57. 

23.  The  roots  of  two  simultaneous  equations  are  represented 
by  the  coordinates  of  the  point  (or  points)  at  which  their 
graphs  intersect. 

24.  Since  two  straight  lines  which  are  not  coincident  nor 
parallel  have  only  one  point  of  intersection,  simultaneous 
linear  equations  have  only  one  pair  of  roots. 

If  two  equations  are  inconsistent,  as2x  +  ?/  —  2  =  0  and  2x  +  y  —  4  =  0, 
their  lines  are  parallel  lines  (§21). 

If  two  equations  are  dependent,  their  graphs  are  identical,  as 

*  +  1  =  l  &n&Sx  +  2y  =  6. 
2      3 

Obviously  inconsistent  and  dependent  equations  cannot  he  used  to  deter- 
mine the  roots  of  a  system  of  equations. 

25.  Equations  of  higher  degree  can  have  several  points  of 
intersection,  and  hence  several  pairs  of  roots. 

Ex.  1.     Solve  graphically  the  following  system : 

x>  +  f  =  25,  (1) 

[3x-2y=-6.  (2) 

Solving  (1)  for  y,  y  =  V25  -  x2. 

Therefore,  if  x  equals  -5,-4,-3,-2,-1,  0,  1,2,  3,  4,  5,  y  equals 
respectively  0,  ±  3,  ±  4,  ±  4.5,  ±  4.9,  ±  5,  ±  4.9,  ±  4.5,  ±  4,  ±  3,  0. 

Locating  the  points  (-5,  0),  (-4, 
+  3),  (_4,  -3),  etc.,  and  joining,  we 
obtain  the  graph  (a  circle)  ABC  of  the 
equation  x2  +  y2  =  25. 

Locating  two  points  of  equation  (2), 
e.g.  (-  2,  0)  and  (0,  3),  and  joining  by  a 
straight  line,  we  obtain  DE,  the  graph  of 
3x-2y=-Q. 

Since  the  two  graphs  meet  in  two 
points  P  and  Q,  there  are  two  pairs  of 
roots,  which  we  find  by  measurement, 
x  s:  1J,  y  =  4$,  or  x  =-  4,  y  =-  3. 


Y 

E 

• 

,?P- 

X 

/ 

/ 

\ 

B 

/ . 

X' 

O 

\x 

- 

■  / 

1 

i 

i 

! 

' 

r  5i' 

-1 

/r 

-2 

[ 

3/ 

\t 

-4 

'  C 

J 

V 

1 

SOLUTION   OF  SIMULTANEOUS  EQUATIONS 


21 


Ex.  2.   Solve  graphically  the  following  system: 


xy 


12, 


(1) 
(2) 


From  (1)  y  = 


12 


x  —y  =  2. 
Hence,  by  substituting  for  x  the  values  —  12,  —  11, 


•  ••  to  +  12,  we  obtain  the  following  points  :  (—  12,  —  1),  (—  11,  —  1^), 
(-  10,  -  1|),  (-  9,  -  H),  (-8,  -li),  (-  7,  -  If),  (-  6,  -  2),  (-  5, 
-2|),  (-4,  -3),  (-3,  -4),  (-2,  -6),  (-1.  -12),  (0,  ±ao),  (1, 
12),  (2,  6),  etc.,  to  (12,  1). 

Locating  these  points  and  joining  them  produces  the  graph  of  (1),  which 
consists  of  two  separate  branches,  CD  and  EF. 

Locating  two  points 
of  equation  (2)  and 
joining  by  a  straight 
line,  we  have  the  graph 
AB  of  the  equation  (2). 

The  coordinates  of 
the  two  points  of  inter- 
section P  and  P'  are  the 
required  roots.  By  act- 
ual measurement  we 
find  x  =  4.5+,  y  =2.5+, 
or  x  =  —  2.5,  y  =  —  4.5. 

To  obtain  a  greater 
degree  of  accuracy,  the 
portion  of  the  diagram 
near  P  is  represented 
on  a  larger  scale  in  the 
small  diagram.  Since 
the  small  part  of  CD 
which  is  represented  is  almost  a  straight  line,  it  is  sufficient  to  locate  two 
or  three  points  of  this  line.     By  actual  measurement  we  find : 

x  -  4.606,  y  =  2.606. 

Evidently  the  second  pair  is 

as  =  —2.606,  y  =  -4.606. 

By  increasing   the   scale   further,  any  degree   of  accuracy  may  be 
obtained. 


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22 


GRAPHIC  ALGEBRA 


EXERCISE  7 


4     Ux  +  3y=12, 
\    x  +  5y=6. 


5. 


3x  +  5y  =  7. 

5  x  —  y  =  7. 


3. 


Solve  graphically  the  following  simultaneous  equations 
■3x  +  4y  =  8, 
2x-3y  =  6. 
3x  +  4y  =  10, 
4  x  +  y  =  9. 

2a;-32/  =  7, 
3z  +  2?/=-8. 

6.  Show  graphically  that  the  following  system  cannot  have 
finite  roots  : 

\2x-y=2, 
[2x-y  =  6. 

7.  Show  graphically  that  the  following  system  is  satisfied  by 
an  infinite  number  of  roots  : 

4  +  3       ' 

3x  +  4y=12. 

8.  Without  constructing  the  graphs,  determine  the  relative 
positions  of  the  loci  of  14  x  —  7  y  -f  2  =  0  and  14  x  —  7  y  +  5  =  0. 

Solve  graphically : 


10. 


11. 


12 


r^  +  2/2=i6, 

# +  2/ =  5, 
./•//  =  6. 

«-y  =  l, 

x2  +  2/2  =  25. 

*  -  y  =  2> 

xy  =  8. 


13. 


14. 


15.    \ 


16. 


(4x-5y  =  10, 
\xy  =  6. 

x2  —  y2  =  4, 


25. 

=  12, 

=  8. 


a-  =  2  y. 

f.r//=  6, 


x2  +  xy-- 
x2  -  y2 : 


26.  The  equation  of  the  circle.  77ie  locus  of  an  equation  of  the 
form  x2  -\-  y2  =  r2  (1)  is  a  CiYcZe  whose  center  is  the  origin  and 
whose  radius  is  r. 


SOLUTION  OF  SIMULTANEOUS  EQUATIONS 


23 


For  the  distance  from  the  origin  0  of  a  point  P  in  the  locus, 

OP=V.?+?  (1) 

=  Vr2  =  r. 

But  if  the  distance  of  every  point 
in  the  locus  from  0  is  equal  to  r,  then 
the  locus  is  a  circle  whose  center  is  0 
and  whose  radius  is  r. 

Thus,  x2  +  y2  —  16  is  a  circle  whose  center 
is  0  and  whose  radius  equals  4,  x2  +  y2  ~  10 
is  a  circle  whose  center  is  0  and  whose  radius  is  VlO. 

Note.  The  square  root  of  a  number  can  often  be  represented  by  the 
hypotenuse  of  a  right  triangle  whose  arms  are  rational  numbers.  Thus, 
VlO  =  V32  +  l2,  hence  VlO,  equals  a  line  joining  (0,  0)  and  (3,  1). 


i 

,r 

P 

u 

X' 

0 

X 

X 

• 

\y' 

27.    The  locus  of  expressions  of  the  form 

(x-ay+(y-by  =  r* 

is  a  circle  whose  center  is  (a,  b)  and  ivhose  radius  eqxials  r. 
Let  P  beany  point  in  the  locus,  and  C=  (a,  h). 


(2) 


Draw  CD  ||  OX ; 


±£^k 


CP  =  CD"  +DP\ 

But  CD  =  x  —  a,  and  DP  =y  —  b. 

.:CP2=(x-a)2+(y-by. 

Hence,  from  (2),  CP2  =  o2,  or  CP=r. 

I.e.  the  distance  of  any  point  iu  the 
locus  from  C  equals  r,  or  the  locus  is  a 
circle  whose  center  is  (a,  b)  and  whose  radius  is  r. 

Thus,  (x  —  2)2  +{x  +  4)2  ^8  represents_a  circle  whose  center  is  (2,  —  4) 
and  whose  radius  equals  V8.  Since  V8  =  V22  +  22,  it  is  easily  con- 
structed. 

Note.  The  equation  (x  —  a)2+  (y  —  b)2  =  r2,  however,  represents  a 
circle  only  if  the  scale  units  of  the  abscissas  and  ordinates  are  equal.  If 
the  two  scales  are  unequal,  the  locus  is  an  ellipse. 


24 


GRAPHIC  ALGEBRA 


Ex.  1.   Construct  the  locus  of 

x2  +  2x+y2-4y-5  =  0. 

Transpose  and  complete  the  squares    of  the  expressions  involving  x 
and  y, 

(x2+2x  +  l)  +  (2/2-42/  +  4)  =  5  +  6, 
(x +1)2+ 0-2)2  =  10. 
I.e.  the  required  locus  is  a  circle  *vhose  center  is  (  —  1,  +2)  and  whose 

radius  is  vTo. 


0. 


3     l       9       l     9 
2  +  TS  T"  ?J 


Ex.  2.  Construct  the  locus  of 

2a;2  +  2?/2-3a;  +  6?/  +  3 
Dividing  by  2,  and  transposing, 

X2_3x  +  2/2  +  32/=_3< 

Completing  the  squares, 

X2_3x+(3)2  +  2/2+3y+(|)2=. 

(^-f)2+(2/  +  l)2  =  fi- 
I.e.  the  locus  is  a  circle  whose  center  is  (f ,  —  |)  and  whose  radius 

IV21. 

28.  The  preceding  examples  show  that  the  locus  of  a  quad- 
ratic function  involving  two  variables  is  a  circle,  if  the  func- 
tion does  not  contain  xy  and  if  the  coefficients  of  x2  and  y2  are 
equal. 

EXERCISE   8 


is 


Solve  graphically : 

*2  +  2/2  =  4, 
x  +  y  =  3. 

x2  +  y2  =  16, 
x  —  y  =  4. 

a;2+2/2  =  50, 
x  —  y  =  —  6. 

fx2  +  2/2  =  9, 
[x  —  2y  =  2. 
x?  +  y2=16, 
2y-3a  =  6. 


6. 


7. 


4. 


5. 


9. 


10. 


[x2-2x  +  y2-±  y  =  0, 

[y  =  2x. 

x2-4:x  +  y2+2y+3=0, 
x  —  y  =  3. 

x2-10x  +  y2  =  0, 
x2  +  6  a;  +  y2  =  16. 

(x-r-l)2-(2/-l)2  =  2, 
(o;-l)2+(2/  +  l)2  =  8. 

'aj2  +  2/2  =  l, 
(a:-l)2  +  y2  =  2. 


PART  II 

SOLUTION    OF   EQUATIONS   BY   MEANS    OF 
STANDARD    CURVES 

29.  A  disadvantage  of  the  preceding  graphic  methods  is  the 
fact  that  they  often  require  a  great  deal  of  numerical  calcula- 
tion, and  that  the  necessary  curves  are  difficult  to  draw.  In 
the  following  chapters,  methods  will  be  given  for  the  solu- 
tion of  quadratics,  cubics,  and  biquadratics  by  means  of  one 
standard  curve,  and  straight  lines  or  circles ;  i.e.  one  curve 
may  be  used  to  solve  all  quadratics  or  all  cubics,  etc.  The 
construction  of  these  curves  requires  very  little  calculation, 
and  once  constructed,  each  curve  may  be  used  for  the  solution 
of  many  problems. 

Three  curves  are  used  in  the  following  chapters,  viz.  a 
parabola    y  =  x2,   a   cubic   parabola   y  =  oc?,    and    an    equilateral 

hyperbola  y  =  -. 

y  =  x2  was  drawn  and  discussed  in  §  8. 

A  locus  of  the  form  y  =  -  was  given  in  §  25,  Ex.  2,  and  the  graph  of 

x 
y-=xz  will  be  given  in  §  49. 

Any  one  of  these  three  curves  may  be  used  to  solve  with 
rules  and  compasses  either  quadratics  or  cubics,  but  only  the 
parabola  and  equilateral  hyperbola  yield  simple  solutions  for 
biquadratics. 

25 


CHAPTER   V 

QUADRATIC  EQUATIONS 

30.   To  solve  the  quadratic 

ax2  +  bx  +  c  =  0  (1) 

by  means  of  a  standard  curve,  we  split  the  equation  (1)  into 
two  simultaneous  equations,  one  of  which  is  the  standard 
curve,  while  the  other  is  a  straight  line  or  circle. 

Thus,  if  ax2  +  bx  +  c  =  0,  (1) 

Let  y  =  x*.  1  (2) 

Substituting  in  (1),       ay  +  bx  +  c  =  0.  J  (3) 

The  solution  of  the  system  (2),  (3)  for  x  produces  the 
required  roots  of  (1). 

x  But  the  graph  of  (3)  is  a  straight  line,  while  the  graph  of 
(2)  is  identical  for  all  quadratic  equations.  Hence,  after  the 
graph  y  =  x2  (see  annexed  diagram)  has  been  constructed,  any 
quadratic  equation  may  be  solved  by  the  Construction  of  a 
straight  line,  provided  the  roots  lie  within  the  limits  of  the 
represented  abscissas  (—6  and  +  6). 

Ex.  1.     Solve  11  x2  +  30  x  - 165  =  0.  (1) 

Let  y  =  x2.  (2) 

Then  11  y  +  30  x  —  165  =  0.  (3) 

In  (3),  if  x  =  0,  then  y  =  15 ;  if  y  =  0,  then  x  =  5 J.  Tbe  straight  line 
joining  the  points  (0,  15)  and  (5|,  0)  is  the  graph  of  (3),  which  intersects 
the  graph  of  (2)  in  P  and  P'.  By  measuring  the  abscissas  of  P  and  P', 
we  have  x  =  2.7,  or  x  =  —  5.5. 

Ex.2.     Solve  5  x?  - 14  x-  65  =  0.  (1) 

Let  y  =  a;2.  (2) 

Then  by- 14* -65  =0.  (3) 

26 


QUADRATIC  EQUATIONS 


27 


Locating  two  points  of  the  equation  (3),  e.g.  (0, 13)  and  (5,  27),  and 
joining  by  a  straight  line  produces  the  graph  of  (3),  which  intersects  the 
graph  of  (2)  in  Q  and  Q'.     Measuring  the  abscissas  of  Q  and  Q',  we 

obtain 

a  =  5.3,  or  x  =  —  2.5. 


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31.    In  the  equation  ay  +  bx  +  c  =  0,  if  x  =  0,  then  y  = , 

and  if  y  =  0,  then  *=  — ?•     Hence,  by  laying  off  on  the  a>axis 

the  distance  —  -  and  on  the  w-axis  the  distance ,  and  apply- 

b  a 

ing  a  straight  edge,  the  roots  of  the  equation  ax2  +  bx  +  c  =  0 

can  frequently  be  determined  by  inspection. 

If  the  two  points  constructed  on  the  axes  lie  very  closely 

together,  the  drawing  is  likely  to  be  inaccurate,  and  it  is  better 

to  locate  one  or  both  points  outside  the  axes. 


28  GRAPHIC  ALGEBRA 

EXERCISE  9 

Solve  the  following  equations  by  the  graphical  method : 

1.  xt-x-6  =  0.  8.   4x2-25«  +  20  =  0. 

2.  ^_|_a;_2  =  0.  9.   3ar  +  20o;  +  12  =  0. 

3.  a2- 3 a -18  =  0.  10.   a^  +  aj  — 5  =  0. 

4.  a?2  +  3aj-10  =  0.  11.  x3  —  2x  —  9  =  0. 

5.  a2-2a;-8  =  0.  12.    3x*  +  7x-42  =  0. 

6.  ^  +  2^-4  =  0.  13.   2^  +  5^-20  =  0. 

7.  ar!_5a;-15  =  0.  14.    5r-4a;-5  =  0. 

32.  Solution  for  large  roots.  By  changing  the  unit  of  the 
abscissas  and  the  unit  of  the  ordinates,  the  same  diagram  may  be 
used  to  represent  y  =  v?  for  various  values  of  x.  For  in  the  dia- 
gram we  may  assign  any  values  to  the  abscissas,  provided  the 
corresponding  ordinates  are  made  equal  to  the  squares  of  the 
abscissas.  Thus  after  the  graph  of  y  =  x2  has  been  drawn  from 
x  =  — 10  to  x  =  10,  we  may  multiply  the  numbers  on  the  «-axis 
by  any  number,  e.g.  3,  and  thereby  extend  the  diagram  from 
x  =  —  30  to  x  =  30,  provided  we  multiply  the  numbers  on  the 
?/-axis  by  32,  or  9. 

This  change  of  scale  units  does  not  affect  the  character  of 
the  locus  ay  +  bx  +  c  =  0,  for  this  equation  is  a  straight  line 
whether  the  abscissas  and  ordinates  are  drawn  on  the  same  or 
different  scales. 

The  annexed  diagram  can  be  used  directly  for  roots  between 
— 10  and  + 10.  If  the  roots  are  larger,  but  lie  between  —  100 
and  +100,  multiply  the  units  by  10  and  102  respectively;  i.e. 
omit  the  decimal  points  in  the  diagram. 

For  roots  still  larger,  add  another  cipher  to  the  values  of 
the  abscissas  and  two  ciphers  to  the  values  of  the  ordinates. 

Ex.  1.    Solve  graphically  x2  — 16  x  -  4400  =  0. 

Let  2/  =  x2; 

then  y-  16  x-  4400  =  0. 


QUADRATIC  EQUATIONS 


29 


Obviously  the  regular  diagram  does  not  contain  the  required  roots. 
Hence  multiply  the  values  of  the  abscissas  by  10  and  the  values  of  the 
ordinates  by  102 ;  i.e.  disregard  the  decimal  points. 


30 


GRAPHIC  ALGEBRA 


C 

b 


Since  ^  is  very  large,  locate  two  points  as  follows 


If  x  =  0,      y  =  4400. 

If  x  =  100,  y  =  6000. 

The  line  joining  (0,  4400)  and  (100,  6000)  intersects  y  =  a?m 


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QUADRATIC  EQUATIONS  31 

P  and  P.     By  measuring  the  abscissas  of  Panel  P,  we  obtain 
x  =  74+  and  x  =  -  59+. 

33.  Small  roots.  Tor  small  roots  multiply  the  values  of  the 
abscissas  by  a  fraction,  most  conveniently  by  .1,  and  the  values 
of  the  ordinates  by  .01 ;  i.e.  place  the  decimal  point  in  front  of 
each  number  given  in  the  diagram  (except  x  =  10  and  y  =  100, 
which  become  1.0  and  1.00  respectively). 

Thus,  1.0,  2.0,  3.0,  etc.,  become  .10,  .20,  .30,  etc.  As  this 
shifting  of  the  decimal  point  is  a  simple  operation,  it  may  be 
done  mentally,  without  any  actual  alterations  of  the  numbers 
in  the  diagrams. 

Ex.  2.    Solve  10  ar>  +  5a?  =  l. 


Let 

y  =  x2. 

Then 

10  y  +  bx  =  1. 

If 

x  =  0, 

y  =  A. 

If 

x=l, 

y  --  A. 

Since  in  the  original  diagram  such  small  fractions  of  y  cannot  be  well 
represented,  multiply  the  numbers  on  the  x-axis  by  .1  and  the  numbers 
on  the  ?/-axis  by  .01;  i.e.  imagine  the  decimal  point  to  be  placed  in  front 
of  each  number. 

Then  the  straight  line  that  joins  (0,  .1)  and  (1,  —.4)  intersects  the 

parabola  in  Q  and  Q'.     The  measurement  of  the  abscissas  of  Q  and  Q' 

gives  the  roots 

x  =  .15,  or  x  =  —  .65. 

Note.  The  student  should  draw  a  diagram  similar  to  the  one  used  in 
the  text,  but  on  a  larger  scale.  The  cross-section  paper  employed  should 
have  each  unit  divided  into  10  parts. 

EXERCISE  10 
Solve  graphically : 

1.  x2-  15 x -4500  =  0.  6.  x2  +  80 x  =  -  700. 

2.  x2  - 10  x  -  3000  =  0.  7.  x2  - 10  x-  600  =  0. 

3.  or  +  80  x  + 1200  =  0.  8.  .^  +  8^-128  =  0. 

4.  x2  +  40a;=1200.  9.  x2  -  30  x-  1800  =  0. 

5.  x2  +  30  x  =  4000.  10.  x2  +  33  x  - 1210  =  0. 


32 


GRAPHIC  ALGEBRA 


11.  2  or2  4- 3  a -1500  =  0.  • 

12.  3  or2 +  10  a;  =  3000. 

13.  ar  + 29  a  =210. 

14.  3x£  +  200  a- 1200  =  0. 

15.  50^-15^-6  =  0. 

16.  10a^-6a;-l=0. 


17.  4af!  +  5a;-l  =  0. 

18.  20  x2  +  3  x  -1  =  0. 

19.  50ar-5.r-3  =  0. 

20.  25 a2 +  10 a- 3  =  0. 

21.  50.c2  +  5a;-l  =  0. 

22.  8  x2  -2  x  — 1  =  0. 


(1) 
(2) 
(3) 


34.    Graphic  representation  of  a  quadratic  function. 
Consider  the  equation 

x2  +px  +  q  =  0. 
Let  y  =  x2. 

Then  y  +px  +  g  =  0, 

or  2/  =  —  pa  —  q. 

In  the  annexed  diagram,  let  COD  represent   the   parabola 
y  —  x2,  and  BR  the  straight  line  y  +  px  +  q  =  Q,ov  y  =  —  px  —  q. 

Let  CL4  or  x'  be  any  particular 
value  of  x, 

then  (L4  =  x'2, 

and  2L1  =  —  px'  —  g. 

Hence 
CB=OA-BA  =  x'2  +  px'  +  g. 

I.e.    the    value    of  the  function 

x2  +  px  +  q    for     any    particular 

value  x'  is  represented  by  that  part 

of  the  corresponding  ordinate  which 

is  intercepted  between   the   straight 

line  y  +  px  +  q  =  0,  and  the  parabola  y  =  x2.     The  distance  is 

measured  from  the  straight  line,  and  is  taken  positive  if  it 

extends  upward,  negative  if  it  extends  downward. 

Thus,  in  the  annexed  diagram, 

y  =      x2  —  \x  —  1 ,  and  we  have : 


\ 

■y 

f 

\ 

O 

<?' 

\ 

4^ 

V 

y* 

3 

& 

^ 

\ 

F  8 

A 

1 

\^- 

\, 

0 

T 

2 

i 

O 

r' 

4 

If 
If 
If 


x  =  -l,y  =  HF=\, 
x=      l,y  =  KI  =  ~l 
x  —      2,  y  =  2,  etc. 


QUADRATIC  EQUATIONS 


33 


Note.  If  we  consider  the  distances  cut  off  from  SB  by  the  ordinates, 
as  abscissas,  e.g.  SK=  x  =  1,  then  the  parabola  represents  the  function 
x2  +  px  +  q  in  so-called  "  oblique  coordinates." 

Ex.  1.  Find  the  values  of  x  which  will  make  the  function 
x2—\x  —  \  equal  to  2,  i.e. 


x2 


\x-l  =  2. 


On  YY'  lay  off  SL  =  2,  and  through  L  draw  PQ  II  BE,  meeting  the 
parabola  in  P  and  Q.     By  measuring  the  abscissas  of  P  and  Q  we  find 

x  ——  |,  or  2. 

Ex.  2.    Find  the  smallest  value  of  the  function  x2  —  2  x  —  1. 

Construct  AB,  the  locus  of  y  —  2x  —  1=0.  (See  next  diagram.) 
Draw  a  tangent  parallel  to  AB,  touching  the  parabola  in  C ;  then 
DC  =  —  2  is  the  required  value. 

35.  To  construct  the  graph  of 
x2  +  px  +  q  in  the  usual  manner 
(rectangular  coordinates),  make 
H'G'  =  HG,  E'F'=EF,  I'K'=  IK, 
etc.  The  curve  G'F'C'L'  is  the 
required  locus  of  x2+px  +  q  (i.e. 
of  x2  —  2a:  — 1). 

Note.  §  35  makes  it  possible  to  con- 
struct the  locus  of  a  quadratic  function 
without  any  computation. 

36.  The  value  of  the  function  ax2 
+  bx  +  c  is  equal  to  a  times  the 
part  of  the  corresponding  ordinate 
ivhich  is  intercepted  by  the  straight 
line  ay  +  bx  +  c  =  0,  and  the  pa- 
rabola y  =  x2. 

For  ax2+bx+c  =a(x2-\-  ■ 

b        c 
But  the  function  x2  H — sc  +  -  is  represented  by  that  part  of 

a        a  , 

the  ordinate  that  lies  between  y  —  x2  and  y  +  -x  +-  =  0  or 

a        a 

ay  +  bx  +  c  =  0.     Hence  ax2  +  bx  +  c  is  equal  to  a  times  this 
intercept. 


-x+c- 
a       a 


34  GRAPHIC  ALGEBRA 

EXERCISE   11 

Find  graphically : 

1.  The  value  of  x2  -  2  x  -  2,  if  x  equals  -  3,  -  2,  .5,  1±. 

2.  The  value  of  x?-\-  x  —  3,  if  a;  equals  — 1.5,  —  1,  0,  2. 

3.  The  value  of  x2  —  5  x  — 12,  if  x  equals  —  2.1,  — 1.5,  3.5„ 

4.  The  value  of  x2  +  4  x  +  5,  if  sc  equals  —  7.5,  9.4,  —  8.8. 

5.  The  values  of  x  if  x2  —  2  x  —  10  =  4. 

6.  The  values  of  x  ii  x2  + 10  x  + 10  =  5. 

7.  The  smallest  value  of  x2  —  4  a;  +  3. 

8.  The  smallest  value  of  x2  + 10  x  —  5. 

9.  The  smallest  value  of  x2  -f  7  a;  —  3. 

10.  The  smallest  value  of  cc2  —  5  x  +  2. 

11.  The  value  of  2  x2  +  6  a  +  7,  if  a;  equals  -  3,  6.5,  7. 

Without  calculating  the  various  values  of  the  function,  con- 
struct the  loci  of :  , 

12.  x2  +  Qx  +  10.  15.   tf  +  hx—'S. 

13.  a^  +  4a;-5.  16.    a2  — 3  a; +  7. 

14.  a;2  —  4  a; +  7.  17.    a'2  +  .c  +  l. 

18.    a;2 +  4  a;  — 7. 

37.  Equal  roots.  If  the  line  ay  +  bx  +  c  =  0  is  a  tangent 
to  the  parabola  y  =  x2,  the  two  points  of  intersection  coincide, 
and  the  two  roots  of  ax2  +  bx  +  c  —  0  are  equal. 

Ex.  1.    Solve  jb2  -  8  x  + 16  =  0. 

Let  2/  =  x'2. 

Then  ?/-8x+l6  =  0. 

If  x  =  0,  ?/ =  -16. 

If  *  =  10,  y  =  G4. 

The  line  RR',  which  joins  (0,  —16)  and  (10,  64),  is  tangent  to  the 
parabola  at  the  point  R. 

Measuring  the  abscissa  of  R,  we  obtain  two  equal  roots,  4  and  4. 


QUADRATIC  EQUATIONS 


35 


=:    __  *s"    „ K  __:__ _ __  __^»     _:___: :  _2__       = 

1                                         *s                                                 1     \                                                                                                                                          J'                                                                                '3 

s^                         \                                                                                                                            / 

=>       _  _  -       :  sr  jl  zlizii  _.  _    -.    .-            ::=i     -  ::  zizz  _-     = 

RK  s.        V9*                                                          \ 

i                                             *■<    V        \*                                                                      T' 

i                              c.;;:z.s3:                           s                '/                <= 

4-               -                    -     I   -   ^v     v?b                                  -?                    x                     X 

\          %     v.P.                                                           - 

±      :         ::              j         \  i^'-       :              :  x    x  :      i    ::       :    i 

=_  _  ::.  :  .  _  :      :; :         ^sri  .    _::  .    :    ::=>    :.:    Izz - c 

^r-                                                                                                        t-                                         ^V*.                                                                            ■■€■                                    7                                                  -             -f 

X                                              t-                   SX        ..                   ...    Z..        '    .      Z.          ; 

X                                      --      i                       55,.                              -          -      2 

l                         ^i                                             / 

C-  --  -x_  -  --  -x:  -X            __  :v  ::       :      :?_    ::  ,*__  _    .: ::= 

a                                                           k                   >?          '                         -# 

-4*                                                                        V-                                                              -            j!                                                                  . 

:              ::::::    j\  .  ±z:         :    :*>         :    :  :         r       :..::       :::  : 

-       jJj           X            -                                               ^r^W-                                                    -1 

oi                                       \SC     ^         =>          r- 

cj|      |                                    V                  "         ' 

.  4,         X       ^S                                 ►                  £                   : 

:                       :       ;  js1  :±:  :       :    :       v              -,__              _         _ 

3-xx                           s              tzz                        z 

i .:      .::  2l  i    ::    .  _    & 2  j..    -    -.       --  _  - 

s    .  :     .                   -£4x    -1 z         :    -r-V-    wztz                           z» 

^ix                            &    x 

ri                                \S    / 

Zp.       ,                                V 

i                          ...       -i^A'X     ._                  _           _ 

-V                                       --           --u=yj--a,        --_--_-------. 

--                                             H                                                      X    'L±X                                                        ^ 

:r __ _  _           _.                        __2_^<=   s-             _.    . 

--* ■-■;               --* \          r 

-----                X              h              -                        r.         -X                \.;,.X.         ..         X 

1           t                         J          :±:        5c.                    x 

---                    X           i                            :.-..  j  ±                 X                      x 

=    .. -  _    _:± i         -_       :i2  -      zl°      -                           x 

X              ~               "X          k       -         X-jX-              -       ::~:± 

x    --------  --.±          t                /            /  ±:    -                           j 

:i                                                                                                              \                               N/                              /         ?i                                                                                        0 

.:.  .  :              .-              j       -    v?x       ~  x    x                          --       i 

-----                        -&/           5      x 

v>^         -X      t  _  _  : 

:                                                   1          IP'           v        °                                        : 

1                                              "  :i'     ::       "                                      * 

X     --                  -                                t       3$v             -■*-          '                                              -I 

X                                  -                          ^S>X       -     z 

x    .                                           i^x    :  /  -                :             -      -- 

=> _      ___           _                             "f     rf~t_    --t    -- 0 ; ;_        'a 

4                                             \    • '         y1                                                u 

'        -                      -                                IX     X-,'                    X 

.-         12,  z       :         x         -- 

i  "       z 

=                                                       /       y                     =>                                       b 

'       ^                       ■--                                        ^ 

/■\      *  \                       1 

/    M 

;                                                                                       '    >f  ^                                                 0                                                                   0 

1     -                                                           -x»"J'-J-p-             -             -             -,              --                                                , 

zf 

sj             \  1 

= ::      _:        sj  _  _  r_         x =                          g 

^                     \ 

=•           .__..            -       ,^-          _               _          X--         --            c-                                       -       - =i 

*J                                        s*\                                                                                     *                                                             *- 

^ 

jS* 

=>                       /^  '                                                                                            0                                                         - 

z,                       ^                                                                                                                        '.-.'■                                       i 

*r           ?i                                fc             *               '"                                 ^               -1               Ej              ,  '■'              '                'L              1 

38.  If  the  roots  are  equal  or  nearly  equal,  the  graphic 
method  is,  however,  liable  to  be  inaccurate,  since  a  slight 
inaccuracy  in  the  construction  of  ay  +  &.«  +  c  =  0  usually 
produces  a  considerable  error  in  the  value  of  x. 


36  GRAPHIC  ALGEBRA 

39.  Complex  roots.      If   the   line   ay  +  bx  +  c  =  0   does   not 

intersect  the  parabola  y  =  x2,  the  roots  are  complex,  and  their 
values  may  be  found  by  means  of  the  following  theorems. 

40.  Let  x2  +  px  +  q  =  0  represent  an  equation  whose  roots  are 
equal;  then  these  roots  are,  by  the  general  formula: 


Hence 

Assuming  that  d  is  a  positive'  quantity,  it  can  easily  be 
shown  that 

x2-\-px-\-q  —  d  =  0  has  the  roots  —  -  ±  y^~.  n\ 

x2  +px  +  q        =  0  has  the  equal  roots  —  -P,  — ^;  (2) 

—  — 

x2  +px  +  q  +  d  =  0  has  the  roots  —  -2  ±  A/_  ci  /$\ 

The  roots  of  (3)  are  complex  and  cannot  be  found  directly 
by  the  graphic  method,  but  if  we  solve  (1)  instead,  we  only 
have  to  multiply  the  irrational  parts  of  the  answer  by  V  —  1 
to  obtain  the  roots  of  (3).  The  straight  line  which  serves  to 
solve  (1)  can  be  obtained  from  the  one  which  solves  (3)  by 
means  of  the  following  proposition. 

41.   If  x2  -\-px  +  q  =  0  has  equal  roots,  the  three  straight  lines 
y+px  +  q-d  =  0,  (la) 

y+px  +  q        =0,  (2  a) 

y+px  +  q  +  d  =  0,  (3  a) 

are  parallel,  and  the  second  one  is  equidistant  from  the  other  two. 
These  lines  (AB,  CD,  and  EF  in  annexed  diagram)  are  parallel  by 
§21. 
By  making  x  =  0,  we  obtain : 

OA  =  -q  +  d, 
OC  =  -q, 

*  Schultze's  Algebra,  p.  269. 


QUADRATIC  EQUATIONS 


37 


OE-  -q-d. 
Hence  AC=CE  =  d; 

I.  e.  CD  is  equidistant  from  AB  and  EF. 

42.  Hence  if  EF  is  known,  draw  the  tangent  CD  ||  EF,make 
CA=EC,    and    construct    AB  \\  EF; 
then  AB   is  the  required   line  whicli 
produces  the  roots  of  (1). 

43.  The  construction,  however,  is 
simplified  by  the  following  theorems: 

1.  TJie  abscissa  of  the  point  of  con- 
tact (Gr)  is  equal  to  the  rational  part  of 
the  roots  of  (1)  and  (3) . 

For  this  abscissa  =  —  "  • 
2 

2.  A   parallel   to    YY'  through    the 
point  of  contact  (G)  bisects  the  chord  KB,  and  hence  any  chord 
parallel  to  the  tangent. 

For  the  abscissas  of  K,  H,  and  B  are  respectively 

OL 


OM-. 


-P-Vd; 

2 

_£• 

'      2' 


2 
Hence  LM  =  MN  =  Vd. 

Therefore,  according  to  a  geometric  theorem,*  KH=  HB. 


To  solve  the  equation 
(1) 


44.    Graphic  solution  for  complex  roots. 

x2  +  bx  +  c  =  0, 

which  has  imaginary  roots. 

Construct  the  locus  EF  of  y  -f-  bx  -f-  c  =  0, 
and  draw  any  chord  PQ  \\  EF.     (See  diagram,  page  38.) 

Through  M,  the  midpoint  of  PQ,  draw  RI  ||  YY'  intersecting 


*  Schultze  and  Sevenoak's  Geometry,  §  144. 


38 


GRAPHIC  ALGEBRA 


the  parabola  in  G,  and  EF  in  J.     Make  GH  =  IG,  and  through 

.H"  draw  i£B  ||  -Ei^7. 

Then  the  abscissa  of  H  is  the 
real  part,  and  the  difference  of 
the  abscissas  of  B  and  H  mul- 
tiplied by  V  — 1  is  the  imagi- 
nary part  of  the  required  roots  ; 
i.e.  x  =  031  ±  MN  x  V  —  1. 

Note.  The  line  RI  may  also  be 
constructed  by  drawing  an  ordinate 

b 


|,  or  if 


through  one  point   [0, 

V  2  al 

o  =  l,  through  (o,  --Y 

45.    If  the  coefficient  of  x2  is 
a,  the  solution  is  the  same  as  if 
this  coefficient  was  unity;   for 
by  dividing  ax2  +  bx  +  c  =  0  by  a,  we  obtain 

aj2+^aj  +  -=0.  (2) 

a        a 

The  straight  line  which  serves  to  solve  (2)  is  therefore 

a        a 
or  a?/  +  6x  +  c  =  0 ; 

t'.e.  we  may  substitute  y  =  x2  directly  into  the  given  equation. 
Ex.  l.   Solves8 -4  a; +  13  =  0. 


Let 
Then 
If 
If 


y  =  %'. 

y-  4x  +  13  =  0. 
x  =  0,  y  =  -  13. 
x  =  5,y=        7. 

Join  (0,  -  13)  and  (5,  7)  by  line  EF, 
and  through  the  midpoint  (R)  of  any 
parallel  chord  draw  RIWYY'.  Make 
GH  =  IG  and  draw  KB  II  EF. 

By  measuring  we  obtain  : 

The  abscissa  of  H  —  2. 


\ 

Y 

'35- 

/l 

\ 

/ 

'b- 

\ 

•    ' 

/ 

\ 

• 

\\/ 

/ 

T 

**.„ 

.  M 

111' 

F, 

;. 

■x 

• 

/ 

A  ' 

K/ 

■>• 

U; 

V 

/ 

A 

>  - 

i 

N 

I    ' 

1 

: 

!  > 

r  - 

1 

P 

(V 

/ 

3 

-If, 

K 

QUADRATIC  EQUATIONS  39 

The  difference  of  the  abscissas  of  B  and  H=  3. 
Hence  the  required  roots  are 

x  =  2  ±  3  V~^l  or  2  ±  3  j. 

Ex.2.    Solve  2  a;2 +5  a;  +  15  =  0.  (1) 

Let  y  —  X2.  (2) 

Then  2  y  +  5  a;  +  15  =  0.  (3) 

Construct  the  line  (3),  i.e.  LN,  and  through  the  midpoint  (M)  of  any 
parallel  chord  draw  MP\]  YY'.  Make  QS  =  PQ  and  draw  SU\\  LN. 
The  roots  produced  by  TUare  -  1.3  ±  2.4.  Hence  the  required  roots  are 
-  1.3  ±2.4V-  1,  or  -  1.3  ±  2.4 i. 

EXERCISE  12 

Solve  the  following  equations  graphically  : 

1.  x2- 10a; +  25  =  0.  8.  a;2  -  5a;  +  15  =  0. 

2.  x2  -  6  x  +  13  =  0.  9.  x2  +  3  x  +  27  =  0. 

3.  a;2 +  4  a; +  8  =  0.  10.  a;2  4- 9  a;  +  36  =  0. 

4.  0^  + 8  a; +  20  =  0.  11.  x2  +  x  +  1  =  0. 

5.  ar9-  8  a; +  25  =  0.  12.  a;2  +  2  a;  +  1  =  0. 

6.  x2  -  10a;  +  29  =  0.  13.  22a;  + 2^  +  3  =  0. 

7.  ar  +  7a;  +  21  =0.  14.  4  a^  -  12  a;  +  25  =  0. 

46.*    Solution  of  quadratic  equations  by  means  of  the  standard 

curve  y  =  -. 
x 

As  stated  in  §  29,"the  parabola  y  =  a;2  is  not  the  only  curve  that 

may  be  used  for  the  graphic  solution  of  quadratic  equations  by 

means  of  straight  lines.     A  curve  that  gives  a  very  convenient 

solution  is  the  equilateral  hyperbola  y  =  -,  which  is  plotted  in 

the  annexed  diagram.    It  consists  of  two  disconnected  branches 
which  approach  the  axes  indefinitely. 

Note.  To  plot  this  curve  exactly,  it  is^necessary  to  locate  several 
points  between  x  =  0  and  x  =  1.  Thus,  if  y'=  2,  x  =  J ;  If  y  =  3,  x  =  I  ; 
if  y  =  4,  x  =  \ ;  etc.     (See  table  on  page  84.) 

*  Paragraphs  marked  by  asterisk  *  may  be  omitted. 


40 


GRAPHIC  ALGEBRA 


47.*   To  solve  the  equation 

ax2  +  bx  +  c  =  0. 


Let 


1  1 

y  =  -,  or  x  =  ~. 
x  y 


(1) 
(2) 


Partly  substituting  this  value  for  x  in  equation  (1), 


Or 


y     y 

ax  +  b  +  cy=0. 
1 


(3) 

(2) 

The  solution  of  the  system  (2),  (3)  for  x  produces  the  re- 
quired roots  of  (1).* 


Ex.  1.    Solve  x2  +  2  x-  8  =  0. 


Y' 

\\ 

O-VfJ 

-TS 

pStf= 

J! 

A 

' 

"NZ-ii 

4—= 

n 

l) 

) 

\     > 

3            •: 

2 

\ 

-1 

\ 

\\ 

> 

V 

■V 

Let 


y  = 


Ex.  2. 
If 


Then        x  +  2-8y  =  0. 

liy  =  0,x=—2;iiy  =  l,x  =  6. 

The  straight  line  that  joins 
(-2,  0)  and  (6,  1)  intersects  (2) 
^  in  P  and  P'.  Measuring  the  ab- 
scissas of  P  and  P',  we  obtain  x  = 
-  4  or  +  2. 

If  the  line    ax  +  b  +  cy  =  0    is 

tangent    to    y  = - . 
equal. 

Solve  4ar*-4a  +  l  =  0. 

1 


the    roots    are 


then 


y 

4x  —  4  +  y 


x 
0. 


(2) 

(3) 


The  line  (3)  touches  (2)  at  Q.    Hence  there  are  two  equal  roots,  \  and  \. 

*  This  method  may  be  used  for  all  equations  of  the  form  ax  f(x)  + 
bf(x)  +  c  =  0. 
Let 

Then 


y  —  —  —' 

ax  +  b  +  cy  =  0. 


QUADRATIC  EQUATION'S 


41 


48.*  Complex  roots  can  be  found  by  a  method  similar  to 
tlie  one  given  in  §  44. 

Students  who  wish  to  derive  this  method  may  be  guided  by  the  follow- 
ing suggestions : 

1.  Consider  the  same  equations 
as  in  §  40. 

2.  These  equations  are  repre- 
sented by  the  lines 

x+p  + (q-d)y -0,       (la) 

x+p  +  qy  =  0,       (2  a) 

x+p  +  (q  +  d)y  =  0.       (3  a) 

3.  Instead  of  being  parallel  (as 
in  §  41)  the  lines  (la),  (2  a),  and 
(3  a)  meet  in  a  point  (B)  on  the 
x-axis  whose  abscissa  is  —p. 

4.  The  lines  (1  a),  (2  a),  (3  a)  intercept  equal  parts  on  any  line  parallel 
to  the  x-axis. 

5.  A  parallel  to  the  y-axis  through  the  midpoint  of    OB  intersects 

y  =  -  at  C,  the  point  of  contact  of  (2  a), 
x 

The  annexed  diagram  solves  the  equation  x2  — x  +  2  =0.  The  line 
x  —  1  +  2y  —  0,  or  BA,  does  not  intersect  the  curve,  but  the  correspond- 
ing line  BB  produces  the  roots  .5  ±  1.3.  Hence  the  required  roots  are 
.5  ±1.3  xV^l. 

EXERCISE  13 

Solve  by  means  of  the  equilateral  hyperbola  the  following 
equations : 

1.  a?  —  2x  —  W  —  0.  4.   35*+ 5a; +  4  =  0. 

2.  x2-x-6  =  0.  5.    a2  — 2»  +  10  =  0. 

3.  ar-6a  +  5  =  0.  6.    ar  +  6a;  +  6  =  0. 

[For  more  examples  see  Exs.  9  and  12.] 

Note.  The  solution  of  quadratics  by  means  of  the  cubic  parabola 
y  —  x3  is  given  in  §  60. 


CHAPTER  VI 
CUBIC   EQUATIONS 

49.  Solution  of  incomplete  cubics.  To  solve  an  incomplete 
cubic  of  the  form  ax3  +  bx  +  c  =  0,  the  method  that  was  used 
for  quadratics  (§  30)  may  be  employed.*     Thus,  to  solve 

ax2  +  bx  +  c  =  0,  (1) 

let  y  =  rf.)  (2) 

Then  ay  +  bx  +  c  =  0.  J  (3) 

The  solution  of  the  system  (2),  (3)  for  x  produces  the  re- 
quired roots. 

But  the  graph  of  (3)  is  a  straight  line,  while  the  graph  of 
(2)  is  a  cubic  parabola  which  is  identical  for  all  cubic  equa- 
tions. Hence  after  the  graph  of  the  cubic  parabola  (AOP  in 
the  diagram)  has  been  constructed,  any  cubic  may  be  solved 
by  the  construction  of  a  straight  line. 

Ex.1.    Solve  4  ^-39  x  +  35  =  0.  (1) 

Let  y  =  x3.  (2) 

Then  4  y  -  39  x  +  35  =  0.  (3) 

In  (3),  if  x  =  0,  then  y  -  —  8|,  and  if  x  =  4,  then  y  =  30 J.     The  line 

joining  (0,  —  8J)  and  (4,  30£)  intersects  the  graph  of  (2)  in  P,  P',  and 

P".     By  measuring  the  abscissas  of  P,  P',  and  P",  we  find  x  =  —  3£,  or 

+  1,  or  2\. 

50.  In  the   equation  ay  -f  bx  +  c,  if  x  =  0,  then  y  =  —  ;    if 

c  a 

y  =  0,  then  x  =  —  r.     Hence,  by  taking  on  the  a>axis  the  point 

G  C 

—r,  on  the  ?/-axis  the  point  — ,  and  applying  a  straight  edge, 
the  roots  of  the  equation  ax3  +  bx  +  c  =  0  can   frequently  be 

*  This  method  may  be  used  for  any  equation  of  the  form  af(x)  +  bx  + 
c  =  0 ;  e.g.  ax5  —  bx  +  c  =  0,  or  x  —  e  sin  x  —  0,  etc. 

42 


CUBIC  EQUATIONS 


43 


determined  by  inspection.  If  the  two  points  thus  constructed 
on  the  axes  lie  very  closely  together,  the  drawing  is  liable  to 
be  inaccurate,  and  it  is  better  to  locate  one  or  both  points  out- 
side the  axes. 

Ex.  2.    Solve  z3  +  6  x- 15  =  0.  (1) 

Let  y  =  %%.  (2) 

Then  y  +  0x-15  =  0.  (3) 

Hence,  the  distances  cut  off  by  (3)  on  the  x-  and  y-axes  are  respec- 
tively 2\  and  15,  and  the  line  (3)  is  easily  constructed.  As  there  is  only 
one  point  of  intersection,  Q,  the  equation  has  only  one  real  root,  viz.  1.7+. 


45 

Y 

A 

*>>  / 

"/ 

^ 

H 

'^o 

P 

<^/G 

-4 

-3 

-2 

-1 

P 

n*^ 

/'SS 

2\ 

3 

4 

X 

X 

V 

II 

„ 

«y 

V 

<& 

p 

/J 

p 

-45 

Y 

44 


GRAPHIC  ALGEBRA 


51.  Solution  for  large  roots.  One  diagram  may  be  used  for 
the  solution  of  large  and  small  roots.  For  in  the  diagram  we 
may  assign  any  values  to  the  abscissas,  provided  the  corre- 
sponding ordinates  are  the  cubes  of  the  abscissas. 

Thus,  after  the  cubic  parabola  y  =  a,*3  has  been  drawn,  we 
may  multiply  the  numbers  on  the  x-axis  by  any  convenient 
number,  e.g.  3,  provided  we  multiply  the  values  of  the  ordi- 
nates by  the  cube  of  the  number,  i.e.  27. 

Similarly,  to  find  small  roots,  mnltiply  the  values  of  the 

abscissas    by    a 
r~  fraction    e.g.   \, 

and  the  values 
of  the  corre- 
sponding ordi- 
nates by  the 
cube  of  this 
fraction,  i.e.  i. 

Ex.  3.  Solve 
graphically  x*+ 
2^-320=0.(1) 

Let  y  =  x3.     (2) 

Then    y  +  2  x  - 

320  =  0.  (3) 

If  x  =  0,  y  =  320, 

and  if  x  =  8,   y  = 

304. 

Obviously  the  preceding  diagram  cannot  contain  the  roots,  and  the 

position  of  (3)  shows  that  there  cannot  be  a  negative  root. 

Hence,  multiply  the  values  of  the  abscissas  in  the  diagram  by  2.  Then 
the  values  of  the  ordinates  must  be  multiplied  by  8.  (The  resulting 
values  are  given  in  parenthesis.) 

Joining  the  points  (0,  320)  and  (8,  304),  we  obtain  the  real  root  6.8", 
■while  the  other  roots  are  imaginary. 

Note.  The  student  should  draw  the  graph  of  y  =  x3  from  x  =  —  3|  to 
x  =  +  31  (or  from  -  4  to  +  4)  on  a  large  scale,  and  use  one  curve  for  the 
solution  of  a  number  of  equations.  The  table  on  page  84  will  be  found 
useful  for  the  construction. 


CUBIC  EQUATIONS 


45 


EXERCISE  14 

Find  graphically  the  real 

roots  of  the  following  equations : 

1. 

a8  +  4  a;  — 16  =  0. 

13. 

aj3-10a?-48  =  0. 

2. 

ar*  -  5  a;  - 12  =  0. 

14. 

£3-9£  +  54  =  0. 

3. 

a?  -  2  x  +  4  =  0. 

15. 

£3-14£  +  24  =  0. 

4. 

2  or5- 9  £  +  27  =  0. 

16. 

tf  _  30  x  - 18  =  0. 

5. 

or3-  7  £  +  6  =  0. 

17. 

or3 +  10  £-13  =  0. 

6. 

4. r3-  39  ^-35  =  0. 

18. 

x3  -  45  x  - 152  =  0. 

7. 

of  _  5  a  +  20  =  0. 

19. 

£3  -  60  x  +  180  =  0. 

8. 

a?  _  5  a  _15  =  0. 

20. 

x?  -  90  x  +  340  =  0. 

9. 

cc3  —  5  x  —  5  =  0. 

21. 

£3-75£-250  =  0. 

10. 

a* -32  a; -80  =  0. 

22. 

£3- 100  £  +  500  =  0. 

11. 

2£3-5£  +  20  =  0. 

23. 

£3  + 120  £-560  =  0. 

12. 

ar5  +  8  x  -  64  =  0. 

24. 

a?  -200  £  +  1200  =  0. 

52.    Graphic  representation  of  a  cubic  function. 

Consider  the  equation 


Let 
Then 


or 


£3+j5£  +  g  =  0. 

y  =  Xs. 
y  +px  +  q  =  0, 

y  =  -px-q. 


/ 

k 

i 

\ 

/3 

7 

L 

5 

K^/ 

&x 

>- 

>-X- 

> 



1 

O 

'^f^ 

L 

S/ 

I 

Sy-- 

z%% 

0 

* 

'A 

G 

{  H 

S^A 

& 

E 

/^i 

s^\ 

& 

7U>      1 

(1) 

(2) 
(3) 


46 


GRAPHIC  ALGEBRA 


In  the  annexed  diagram,  let  COD  represent  the  cubic  pa- 
rabola y  =  x3,  and  BE  the  straight  line  y  -{-  px  +  q  =  0,  or 
y  =  —px—q. 

Let  OA  or  x'  be  any  particular  value  of  x. 

Then  CA  =  x'3, 

and  BA  =  —px'  —  q. 

Hence  CB  =  CA- BA  =  x's+px' +  q. 

I.e.  the  value  of  the  function  x3  -{-px  +  q  for  any  particular 
value  x'  is  represented  by  that  part  of  the  corresponding  ordinate 
which  is  intercepted  between  the  straight  line  y+px  +  q  =  0,  and 


i 

v 

(if 

'B 

•3tpx-hg 

7 

L 

'13 

c 

K^ 

^X 

>- 

P-XJr<Z- 

>-x — 



> 

i 

0 

L 

^i_ 

7lz 

0 

V, 
X 

'A 

G 

(    H 

^jij 

& 

E 

/^ 

y^ 

*y 

D      | 

tfie  cm&i'c  parabola  y  =  a3.     The  distance  is  measured  from  the 

straight   line,   and   is   taken   positive   if   it   extends  upward, 

negative  if  it  extends  downward. 

Thus   in  the   annexed    diagram    y  =  x3  —  Qx  +  %,   and   we 

have 

if  x  =  -2,y  =  FG  =  5, 

x  =  -l\,y  =  HI=l, 

x  =  l%,y=  KL  =  —  2,  etc. 

Ex.  1.   Find  the  greatest  value  of  the  function  a;3  —  7x  +  6, 

for  a  negative  x. 


CUBIC  EQUATIONS 


47 


Construct  AB,  the  locus  of  y  —  7  x+  6  =0.  Draw  CD  parallel  to  AB, 
touching  the  cubic  parabola  in  E ;  then  FE,  or  14,  is  the  required  value. 

Ex.  2.  Which  values  of  x  will  make  the  function  Xs  —  7  x  +  6 
equal  to  4,  ie. 

3^—735  +  6=4? 

On  any  ordinate,  from  the  straight  line  AB,  lay  off  4  units  upward,  as 
JFT^.  Through  G  draw  HI  parallel  to  AB,  intersecting  the  cubic  parabola 
in  P,  P',  and  P".  By  measuring  the  abscissas  of  P,  P',  and  P",  we 
find  x  =  —  2f ,  or  J,  or  2J. 


Note.  If  we  consider  the  distances  cut  off  from  SB  by  the  ordinates, 
as  abscissas,  e.g.  ST=1%,  then  the  cubic  parabola  represents  the  function 
x3  +px  +  q  in  so-called  "  oblique  coordinates." 


53.  To  construct  the  graph  of  x?+px  +  q  in  the  usual 
manner  (rectangular  coordinates),  make  K'L'=KL,  M'N' 
=  MN,  0'B'=OR,  etc.  The  curve  L'N'B'  is  the  required 
graph  of  Xs  +px  +  q. 


48  GRAPHIC  ALGEBRA 

54.  Tlie  value  of  the  function  ax3  +  bx  +  c  is  equal  to  a  times 
the  part  of  the  corresponding  ordinate  which  is  intercepted  by  the 
straight  line  ay  +  bx  +  c  —  0,  and  the  cubic  parabola  y  =  x3. 

The  proof  is  similar  to  that  of  §  36. 

EXERCISE   15 
Find  graphically : 

1.  The    value    of    a3  +  4  a;  — 16,   if   x   equals    —3,  —2.5, 
-2.1,  3.5. 

2.  The  value  of  x3  +  4  x  —  8,  if  x  equals  — 1.6,  —  1.5,  2,  1.5. 

3.  The  value  of  xs  -  6  x  - 15,  if  x  =  -  3,  -  2,  1.5,  3.5. 

4.  The  value  of  x3  -  5  x  +  18,  if  a?  =  -  8,  -  5,  +3,  +1. 

5.  The  value  of  x,  if  x3  —  5  x  — 12  =  5. 

6.  The  value  of  a-,  if  x3  -  5  a;  - 12  =  - 10. 

7.  The  value  of  a?,  if  x3-  5  a;  -12  =  -40. 

8.  The  value  of  x,  if  a,-3  -  5  x  — 12  =  10. 

9.  The  smallest  value  of  x3  —  5  a;  — 12  for  a  positive  a?. 

10.  The  greatest  value  of  ar3  —  5  x  + 10  for  a  negative  a\ 

11.  Construct  the  graph  of  ar3  — 12  x  —  30  =  0. 

12.  Construct  the  graph  of  x3  —  8  =  0. 

Find : 

13.  The  value  of  2  x3  +  9  a;  +  20  =  0,  if  x  equals  3,  2.5,  - 1.5. 

14.  The  value  of  3  x3  +  9  x  —  25  =  0,  if  x  equals  —3,-5,-2. 

15.  The  smallest  value  of  3  x3  —  9  a*  —25,  for  a  positive  a,\ 

55.  The  preceding  paragraphs  may  be  used  to  locate  the 
line  ay  -\-bx-{-c  =  0  by  determining  two  values  of  the  function 
ax3  -f  bx  +  c.  In  applying  this  method  it  is  advisable  to 
reduce  the  coefficient  of  a;3  to  unity  by  dividing  by  a. 

E.g.,  let  2ari-12a,-  +  3  =  0. 

Dividing  by  2,  a;3  —  6  x  +  f  =  0. 

If  a;  =  3,  a3-6a;  +  !  =  10 J. 

If  ic  =  -3,  flJ»_6a;  +  4  =  — 74. 


CUBIC  EQUATIONS 


49 


Through  the  point  A  (whose  abscissa  =  3)  draw  an  ordinate 
meeting  the  cubic  parabola  in  B,  and  on  BA  lay  off  downward 
jBC=101  Similarly,  through 
the  point  E  (whose  abscissa 
=  —3)  draw  a  perpendicular 
EF  upward  equal  to  1\\  join 
FC,  which  is  the  required 
line. 

56.  Equal  roots.  If  the 
line  ay  +  bx  +  c  =  0  is  tan- 
gent to  the  cubic  parabola, 
two  points  of  intersection 
coincide,  and  two  roots  of 
the  equation  ax3  +  bx  +  c  =  0 
are  equal. 

The  straight  line  must  intersect  the  parabola  at  least  once  ;  hence  every 
cubic  equation  has  at  least  one  real  root. 

57.  It  can  be  proved  that  the  sum  of  the  roots  of  an  in- 
complete equation  of  the  form  axs  +  bx  +  c  =  0  is  equal  to  zero. 
Hence  if  one  root  is  m,  and  the  other  two  are  equal,  then  these 

equal  roots   are   each  —  ^;  i.e.  if  the  abscissa  of  P=m,  the 

abscissa  of  C,  the  point  of  contact,  equals  —  — . 

Since  it  is  difficult  to  locate 
graphically  a  point  of  contact  with 
accuracy,  it  is  advisable  to  deter- 
mine equal  roots  by  the  preceding 
relation. 

58.  Complex  roots  of  incomplete 
cubics;  If  the  line  ay  +  &.r  +  c  =  0 
meets  the  cubic  parabola  in  only  one 
point,  then  two  roots  are  complex.  To  find  complex  roots 
of  the  form  n  ±  V^i,  we   employ  the  same  method  as  for 


50 


GRAPHIC  ALGEBRA 


quadratic  equations ;  viz.  we  determine  the  line  that  produces 
the  roots  n  ±  V  £. 

If  the  equation  ax3  +  bx  +  c  =  0  has  one  root  equal  to  m,  the 
left  member  is  divisible  by  x  —  m,  and  the  equation  may  be 
represented  in  the  form 

a(x  —  m)(x2+px-{-q)  =  0. 
Supposing  that  x2  +  px  +  q  =  0  has  equal  roots,  and  that  d  is 
a  positive  quantity,  we  consider  the  equations  : 

a(x  —  m)(x2  +px+q—  d)  =  0,  (1) 

a(  x  —  m)(x2 -{- px  +  q)         =0,  (2) 

a(x  —  m^x2  +  px  +  q  +  d)  =0.  .   (3) 

In  the  same  manner  as  in  §  40  it  follows  that  the  roots  of 
the  equations  (1),  (2),  and  (3)  are  respectively 

m,  - 1  ±  Vd ; 

P      V 

m,  —-,  -1-  ; 

'      2'      2' 


m, 


V 


±  V^d. 


Hence  the  roots  of  (3)  can 
be  found  by  solving  (1). 

But  the  three  straight  lines 
(la),  (2°),  and  (3a)  which  serve 
to  solve  (1),  (2),  and  (3)  re- 
spectively are  connected  by 
the  following  geometric  rela- 
tions : 

1.  TJie  three  lines  (1°),  (2°), 
and  (3°)  meet  in  a  point  (m, 
m3),  i.e.  P. 

For  m  is  a  root  of  the  three  equa- 
tions (1),  (2),  and  (3). 

2.    Tlie  three  lines  intercept  equal  parts  on  an  ordinate  drawn 
through  the  point  of  contact  C,  or  DC—  CE. 


CUBIC  EQUATIONS 


51 


2 


For  according  to  §  52,  CD  is  equal  to  the  value  of  (3)  if  x  = 


mrt 


Similarly,  it  follows  from  (1)  that  CE  =  -^;  i.e.  CD  and  OE  are  equal 
and  lie  on  opposite  sides  of  C. 

3.  77ie  line  (2a)  is  tangent  to  the  cubic  parabola  at  C. 
This  follows  from  §  56. 

4.  77*e  abscissas  of  D,  C,  and  E  are  equal  to  —  —,  hence  EF 
=  |FP(§57). 

59.    Construction  of   complex  roots. 

Let  ax3  +  bx  +  c  =  0  have  two  com- 
plex roots. 

Substitute  y  =  x3. 

Then  ay  +  bx  +  c  =  0.  (3) 

Construct  PF,  the  locus  of  (3),  and 
let  it  meet  the  parabola  in  one  point, 
P,  and  the  ?/-axis  in  F.  Produce 
PF  by  one  half  its  length  to  E,  and 
through  E  draw  an  ordinate,  meeting  the  cubic  parabola  in  C. 
Produce  EC  by  its  own  length  to  D  and  draw  PD,  intersecting 

the  curve  in  Q  and  R. 
Then  the  abscissa  of  D 
is  the  real  part,  and 
the  difference  of  the 
abscissas  of  Q  and  D  is 
the  imaginary  part  of 
the  required  roots  ;  i.e. 
x=OG±GSV^l. 
Ex.  1.  Solve 
a?  +  x  - 10  =  0.     (1) 

Let         y  =  x3.  (2) 

Then  y+x-  10  =  0.    (3) 
Construct    the  locus  of    (3),  i.e.   FF,   which    intersects    the    cubic 
parabola  in  one  point,  viz.  P. 


52 


GRAPHIC  ALGEBRA 


Hence  the  equation  has  one  real  root,    which    equals    2,  and    two 

imaginary  roots. 

Produce  PF  by  one  half 
its  length  to  E.  Through 
E  draw  an  ordinate  which 
meets  the  curve  in  C. 
Produce  EC  by  its  own 
length  to  Z>,  and  draw 
PD,  intersecting  the  cubic 
parabola  in  Q  and  R. 

The  abscissa  of  Z>=  —  1, 
the  difference  of  the  ab- 
scissas of  Q  and  D  =  2. 

Hence  the  complex 
roots  are  —  1  ±  2  V— 1. 


1 

F 

Y 

10 

1 

X 

I 

5 

% 

X 

3 

- 

9 

c 

. 

!! 

< 

-o 

/ 

} 

-10 

^lS 

^JO-1 

I 

R 

• 

-25 
T 

EXERCISE   16 


Find  the  real  and  complex  roots  of  the  following  equations 


1.  xz~ 3 x -2  =  0. 

2.  a*-3o;  +  2  =  0. 

3.  tf  +  x  _|_  io  _  0. 

4.  4x-3-lla-10  =  0. 

5.  4^-3^-26  =  0. 

6.  x"  +  9  a;  +  26  =  0. 


7.  x3 -9x  +28  =  0. 

8.  a?  —  9x+  280  =  0. 

9.  8ar3-12a  +  9  =  0. 

10.  4^-9^-14  =  0. 

11.  a!*  +  4aj  —  5  =  0. 

12.  a:3  +  2a,-+6  =  0. 


60.*    Solution  of  quadratics  by  means  of  cubic  parabolas. 

To  solve  the  quadratic 

x2+px  +  q=0  (1) 

by  means  of  a  cubic  parabola,  mul- 
tiply by  x—  p,  i.e.  introduce  the  new 
root  p,  x3  +  (q  —  p-)  x  —  pq  —  0.    (2) 

Or  if  y  =  x*,  (3) 

y+(q-p2)x-pq=0.    (4) 

The     line     (4)     passes    through 
(p,  ps)  and  (0,  pq). 

Thus,  to  solve  a;2  +  4  a;  +  3  =  0, 


Y' 

P 

*, 

uu 

1U 

X 

<S  < 

J 

X 

' 

3 

¥^~ 

1 

: 

j     : 

i      ■ 

I    I 

1 

r. 

-20 

^40- 

CUBIC  EQUATIONS 


53 


take  in  the  cubic  parabola  a  point  P  whose  abscissa  =$>  =4,  and  on  OT 
layoff  OB  =pq  =  12. 

The  line  PB  determines  the  roots  (P'  and  P"). 

x  =  —  1  or  —  3. 

Note.     For  examples  see  Exercise  9. 

61.  Complete  cubic  equations.  To  determine  a  method  for 
the  graphic  solution  of  complete  cubic  equations,  consider  first 
a  concrete  example. 

To  solve  or5  +  9  a?  +  20  x  + 12  =  0.  (1) 

Substitute  x  =  z  —  (i  X  second  coefficient). 

Or  x  =  z  -  3.  (2) 

Then  0-3)3  +  9  (2  -3)2  +  20  (2-  3)  +12  =  0.  (3) 

If  (3)  were  simplified,  it  would  not  contain  the  second  power 
of  z,  for  the  first  term  produces  —  9  z2,  the  second  term  +  9  z2, 
and  the  other  terms  do  not  contain  z2. 

Hence  equation  (3)  can  be  solved  by  one  of  the  methods  for 
incomplete  cubic  equations,  but  the  one  given  in  §  55  is  the 
more  convenient,  since  it  does  not  require  the  simplification  of 
the  equation. 

If  z  =  3,  z  -  3  =  0, 

and  (2_3)3  +  9(z-3)2  +  20(2-3)  +  12  =  12. 

If  2  =  -l,z-3==-4, 

and  (z_3)3  +  9(2-3)2  +  20(z-3) +12  =  12. 

Consider  z  as  abscissa,  y  as  ordinate,  and  construct  the  cubic 
parabola  y  =  zs. 

Through  (3,  0)  and  (-1,  0) 
draw  ordinates  and  let  them  meet 
the  curve  in  A  and  C.  On  the 
ordinates  lay  off  downward  AB 
=  12,  and  CD  =  12,  and  draw  BD. 

By  measuring  the  abscissas  of 

the   points   of    intersection,   we 

obtain  the  roots  : 

z=     2,      1, 

Hence  x  =  —  1,  —2, 


and 
and 


3. 
6. 


54 


GRAPHIC  ALGEBRA 


62.  In  the  preceding  diagram  z  represents  the  abscissas,  but 
by  changing  the  location  of  the  ?/-axis  we  can  obtain  abscissas 
which  equal  x. 

On  OX  lay  off  00'  =  3.     Consider  0'  as  the  new  origin, 

and  the  ordinate  Y0Y0',  drawn 
through  0',  as  the  new  ?/-axis. 
Then  the  abscissa  of  any  point 
is  smaller  by  3  than  the  old  ab- 
scissa z,  or  the  new  abscissa  is 
2  —  3,  i.e.  x.  By  thus  introduc- 
ing a  new  axis,  the  entire  work 
of  the  preceding  paragraph  can 
be  done  without  introducing  z  at  all. 

Thus,  instead  of  saying : 

If  z  -  3  =  -  4,  (z  -  3)3  +  9  (2  -  3)2  +  20  (z-  3)  +  12  =  12, 
we  have  briefly : 

If  *=-4,  ar3  +  9a;2+20a  +  12  =  12. 

Similarly,  instead  of  measuring  the  z,  we  may  directly  meas- 
ure the  x,  and  thus  obtain  the  roots  of  (1). 

63.  A  change  in  the  position  of  the  axes  is  called  a  trans- 
formation of  coordinates. 

To  solve  x3  -+-  bx2  +  ex  +  cZ  =  0,  (4) 

we  locate  0',  the  new  origin,  at  the  point  ( ^,  0  V  and  consider 

the  ordinate  through  0'  as  the  new  y-axis.     If  z  is   the   old 
abscissa,  then  the  new  abscissa  x  =  z—  ^,  and  this  value  sub- 
stituted in  (4)  produces  an  equation  without  z2. 
Similarly,  to  solve 

aa?  +  bx2  +  ex  +  d  =  0, 

_6 
3a* 


make 


00' 


64.    The  method  for  solving  complete  cubics,  which  was  derived 
in  the  preceding  paragraphs,  may  be  summarized  as  follows  : 


CUBIC  EQUATIONS 


55 


To  solve  the  complete  cubic, 

ace3  +  bx2  +  ex  +  d  =  0, 


divide  by  a : 


or  +  -or  -j —  x+  -  =  0. 
a  a        a 


Construct  the  standard  cubic  parabola,  and  a/Yer  ££  *'s  cow- 

structed  change  the  origin  to  the  point  ( — ,  0  ]. 

\3a     ) 

Locate  two  points  by  the  method  of  §  55.     The  line  which 

joins  these  points  intersects  the  cubic  parabola  in  one  or  more 

points  whose  abscissas  are  the  required  roots. 

Note.     In  finding  real  roots,  all  work  except  the  construction  of  the 
cubic  parabola  refers  to  the  new  y-axis,  and  the  old  axis  may  be  omitted. 

Ex.1.    Solve  2  x3  - 15  ar  +  31  x  -  12  =  0. 
Dividing  by  2,  and  denotiug  the  left  member  by  y,  we  have 

2x3-  15x2  +  31x-12 


y  = 


2 


=  0. 


After  drawing  the  standard  cubic  parabola  {i.e.  y  =  z3),  lay  off  on  the 
x-axis    00'  =  |  (— -V5-),    i-e- 
—  2\,    and    consider    0'  as 
the  new  origin. 

If        a5  =  0,  y  =  -6. 

If        x  =  2,y=      3. 

Let  the  new  y-axis  (i.e. 
Yo  To')  meet  the  cubic  parab- 
ola in  A,  and  the  ordinate 
through  (2,  0)  meet  the 
curve  in  C.  On  AO'  lay 
off  upward  AB  =  6,  and  on  tffe  ordinate  through  C  lay  off  downward 
CD  =  3.  Draw  BD  and  measure  the  abscissas  of  the  points  of  inter- 
section P,  P',  and  P".    Thus  we  obtain  : 

x  =  \,  3,  and  4. 

65.   Complex    roots    of    complete    cubics   are   determined   by 
applying  §§59  and  64.     In  using  §  59  we  find   the  ordinate 


56 


GBAPHIC  ALGEBRA 


through  the  point  of  contact  by  producing  the  line  PF  from  P 
to  the  ?/-axis  by  one  half  its  own  length.  The  student  should 
bear  in  mind  that  this  refers  to  the  old  y-axis,  or  that  F  lies 
in  TY'. 

Ex.2.     Solve  4  «3  + 18  a2 +  24  a -17  =  0. 
4x3  +  18x2  +  24x-17 


Dividing  by  4,       y  = 


:0. 


2/=-8f. 


Construct  the  cubic  parabola,  lay  off  on  the  x-axis  00'  =  %  of  *£-,  i.e.  1J, 
and  consider  0'  the  new  origin. 

If  x  =  0, 

If  K  =  -3,. 

Locate  the  points  .4 
and  ^4'  in  the  usual 
manner  (AB  =  —  4J, 
A'B'=-8%),  and  draw 
^1^1',  which  meets  the 
cubic  parabola  in  P  and 
the  old  y-axis  in  F. 
Produce  PF  by  one 
half  its  own  length  to 
E,  and  let  the  ordinate 
through  E  meet  the 
curve  in  C.  Produce 
EC  by  its  own  length 
to  Z>,  and  draw  PD  meeting  the  cubic  parabola  in  P'  and  P". 

The  abscissa  of  D  is  —  f ,  and  the  difference  of  the  abscissas  of  P'  and  D 


is 


Hence  the  required  roots  are 

_  |  +  3  v^i,  _  |  -|  V^T,  and 


EXERCISE  17 

Find  graphically  the  real  roots  of  the  following  equations :  * 
1.   x3-3x2-x  +  S  =  0.  3.    z3-6x2  +  3a;  +  10  =  0. 


2.   x3  -9  z2  +  23x- 15  =  0. 


XT 


8  a2  4- 17  a -10  =  0. 


*For  most  of  the  following  examples,  a  graph  of  the  cubic  parabola 
from  x  =  —  3 J  to  x  =  3\  is  sufficient.  In  other  cases,  apply  the  method 
of  §  51. 


CUBIC  EQUATIONS  57 

5.  ar* +7  a;2 +  14 a +  8  =  0.  12.  5ar5-3a;2-20a:+12=0. 

6.  a3-2a:2-5a;  +  6  =  0.  13.  2ar5-4a;2-  10x  +  9=0. 

7.  x3-2x2-4:x  +  2  =  0.  14.  2ar3-5ar2-4a;  +  3  =  0. 

8.  a*-3xi  +  x+7  =  0.  15.  4x3-12»2-19x+12=0. 

9.  ar5  +  4arJ-2x-5  =  0.  16.  4ar5-12a;2-31a;+18=0. 

10.  a?  +  x2  +  x  +  5  =  0.  17.   ^+6 a;2 -24 x  +  60=0. 

11.  2a;3  +  8a;2  +  2a;-3  =  0. 

Find  the  real  and  complex  roots  of  the  following  equations : 

18.  a?-3xi  +  x  +  5=0.  22.  a;3 -6 x2  +  11  a; -12  =  0. 

19.  a^  +  6a;2  +  10a:  +  8  =  0.  23.  x3  +  x2 - 2x  + 12  =  0. 

20.  a3-3ar7  +  2a;  +  6  =  0.  24.  ar3  +  ar-7z  +  15  =  0. 

21.  a;3  +  6a;2  +  13a:  +  20  =  0.  25.  x3-9ar  +  28x-20  =  0. 

66.  Values  of  a  complete  cubic  function.  The  method  for 
finding  the  values  of  a  function  for  various  values  of  x,  as 
given  in  §  52,  is  true  also  for  the  complete  cubic  equation. 

Thus,  in  the  example  of  §  65 : 

If      x  =  -3,  4a;3  +  18a;2  +  24a-17  =  4(yl'B')  =  -35. 

If      x  =  1,  4  ar5  + 18  a;2  +  24a;  - 17  =  4  (IK)  =  29,  etc. 

Note.  In  order  to  make  the  new  y-axis  coincide  with  one  of  the 
lines  of  the  cross-section  paper,  it  is  sometimes  advisable  to  take  the 
unit  of  the  abscissas  equal  to  the  length  of  three  squares  of  the  paper. 

67.  Construction  of  the  graph  of  a  complete  cubic  function. 

f  Ex.  3.    Construct  the  graph  of 

y  =  xs  +  4:X2  —  x  —  4. 

00'  =  |-4  =  f. 

Take  the  unit  of  abscissas  equal  to  the  length  of  three  squares 
(Note,  §66). 


58 


GRAPHIC  ALGEBRA 


Construct  the  cubic  parabola,  and  place  the  new  origin  at  the  point  OK 

If  x  =  0,  y  =  -  4. 

If  x=-2,y  =  6. 

Locate  the  points  A  and  B  in  the  usual  manner,  and  draw  AB.  Draw 
a  new  x-axis  X0X0',  and  make  CD'  =  CD,  E'F>  =  EF,  G'W  =  GH,  etc. 

By  joining  the  points  Z>',  F',  N',  etc.,  in  succession  the  required 
graph  IF'KL  is  obtained. 


68.  If  the  coefficient  of  x3  is  a,  the  student  should  keep  in 
rnind  that  in  applying  the  above  method  every  ordinate  has 
to  be  multiplied  by  a. 

EXERCISE  18 

1.    If    y  =  x3  —  4  ar  +  2  x  +  5,    determine    graphically    the 

value  of  y  if 

(a)  x  =  i  (6)  x  =  If,  (c)  x  =  2. 

Construct,  by  means  of  the  standard  curve,  the  graphs  of  the 
following  functions  in  rectangular  coordinates  : 

2.  y  =  xi  +  2a?-5x-l.  6.   y  =  x3  +  6a;2  -  x  -  30. 

3.  y  =  x3  +  5  x2  —  3.  7.   y  =  x3  +  x2  —  x  +  15. 

4.  ?/  =  aj*+  4  x2  +  x  +  2.  8.   ?/  =  ar3  —  3  x*2  +  7  a;  +  5. 

5.  y  =  4  ar3  -  12  x2  -  19  x  +  12.      9.   ?/  =  x*  +  6  x2  +  2  a-  -  9. 

Note.  The  solution  of  a  cubic  equation  by  means  of  a  parabola  or  a 
rectangular  hyperbola  is  given  on  §§  75  and  84. 


CHAPTER   VII 


BIQUADRATIC  EQUATIONS 

69.    Solution  of  biquadratics  in  which  the  second  term  is  want- 
ing.    To  solve  an  incomplete  biquadratic  of  the  form 

x*  +  bx2  +  ex  +  d  =  0,  (1) 

write  this  equation  as  follows : 

xA  +  (6  -  1)  x2  +  x2  +  ex  +  d  =  0. 
Let  y  =  ^\  (2) 

Then  f  +  (6  -  ±)y  +  sc2  +  ex  +  cZ  =  0.  f  (3) 

The  solution  of  the  system  (2),  (3)  for  x  produces  the  re- 
quired roots.  But  the  graph  of  (2)  is  a  parabola  which  is 
identical  for  all  biquadratic  equa- 
tions, while  the  graph  of  (3)  is  a 
circle  (§  27). 

Ex.  1.  Solve  x*  -  15  x2  -  10  x  + 
24  =  0.  (1) 

Separate  —  15  x2  into  two  parts,  one  of 
which  is  x2 : 

x*  -  16  x2  +  x2  -  lOx  +  24  =  0. 

Let  2/=x2.  W2) 

Then  ?/2-16y+x2-10x+24=0.  J  (3) 

To  construct  (3)  transpose  24  and  com- 
plete the  squares, 
y2-  16  y  +  64  +  x2  -  10  x  +  25 

=  -  24  +64  +  25. 

Or  (2/_8)2  +  (x-5)2=(V65)2.  (3) 
I.e.  (3)  is  a  circle  whose  center  C  is  the 
point  (5, 8)  and  whose  radius  equals  V65.* 


1 

- 

? 

P' 

1 

iO' 

\ 

it 

\ 

\ 

(p 

It) 

J 

\ 

o 

c 

/ 

i 

/ 

/ 

1 

P' 

t 

V 

(P 

XK 

r 

f 

i       i 

ri  < 

LA    2    3    4    5 

£ALj 1 1 1 L J 

*  V65  =  V82  +  l2,  hence  the  line  joining  C  and  (4,  0)  is  the  radius. 
In  other  cases,  use  table  of  square  roots,  Appendix  III. 

59 


60 


GRAPHIC  ALGEBRA 


Equation  (2)  is  the  standard  parabola,  which  is  intersected  by  the 
circle  in  four  points,  P,  P,  P",  and  P".  The  abscissas  of  P,  P,  P", 
and  P'"  are  the  required  roots.     .-.  x  =  —  3,  —  2,  1,  or  4. 

Note.  The  student  should  remember  that  in  problems  involving  cir- 
cles, the  same  scale  unit  must  be  used  for  abscissas  and  ordinates. 

70.  Formulae  for  radius  and  origin.  According  to  the  preced- 
ing paragraph,  the  equation 

a4  +  bx2  +  ex  +  d  =  0  (1) 

is  solved  by  the  system 

y  =  «?,\  (2) 

v?  +  cx  +  f+(b-l)y  +  d  =  0.  J  (3) 

Transposing  and  completing  the  squares  in  (3), 

2 


^  +  «+(5Y+^+(&-i)y+^Y«(|Y+feiY-d 


Or 


r+i' 


2 
&-1V 


+ 


2 


-d. 


IT 

> 

iT" 

P 

< 

JT 

'  1 

r 

14^ 

TO" 

i* 

\ 

[p 

1U" 

' 

/ 

c 

/ 

< 

/ 

o 

/ 

1 

/' 

N 

,Q 

V' 

c 

//' 

■ 

z 

$ 

-■ji  -rrr^i  < 

LA    2    3    4    5 

If  we  denote  the  coordinates  of 
the  center  of  the  circle  by  x0  and  y0, 
and  the  radius  by  r,  we  have 


c 

9> 


(4) 

?/o  =  :Hp>  (-5) 

r2  =  a-024-7/o2-d.       (6) 

Ex.   2.    Solve   by   means   of  the 
formulae : 


aj» 


3aj2  +  4a?  +  3  =  0. 


a;0  —  —  2, 

r2  =       5. 
/.e.  the  center  C"  of  the  circle  is  (  —  2,  2), 
and  its  radius   is   V5,  or  the  line  that 
joins  C  and  (—  1,  0). 


BIQUADRATIC  EQUATIONS 


61 


There  are  only  two  points  of  intersection,  and  hence  two  roots  are  real, 
and  two  complex. 

The  real  roots  are  -  .6  and—  2.1. 


71.*  The  expression  Vx2  +  y2  —  d  can  be  con- 
structed geometrically.  If  C  is  the  center  of  the 
circle,  lay  off  on  the  x-axis  OD  =  0(7,  and  draw  the 
ordinate  DE,  which  equals  x2  +  y2.  On  ED  lay  off 
EF  =  d  and  draw  FH  ||  XO.  The  segment  HG  in- 
tercepted on  this  parallel  by  the  y-axis  and  the  "* 
parabola,  equals  r. 

EXERCISE  19 


Find  the  real  roots  of  the  following  equations :  * 


1.  z4  +  5a2  +  4;c-28  =  0. 

2.  a;4-15ar  +  10a;  +  24  =  0. 

3.  xA  —  x2  +  4  x  —  4  =  0. 

4.  3^-19^  +  2  x  +  56  =  0. 

5.  x4-5r  +  4=0. 

6.  a4  -7  a;2  -12  a +  18=0. 


7.  a;4  — 4a?  +  12ar  +  9=0. 

8.  a4-7ar-6a;  +  12  =  0. 

9.  z4-  9  «2  -2^  +  6  =  0. 

10.  a;4  +  4  x2  —  5  x  —  55  =  0. 

11.  x-4- 6^  +  30;  + 2  =  0. 

12.  x4  -  15a2-  10x  +  24  =  0. 


72.    Solution  for  large  roots.      To  use  the  same  diagram  of 

the  standard  curve  for  the  finding  of  large  and  small  roots  of 

the  equation 

x*  +  bx2  +  ex  +  d  =  0,  (1) 

multiply  the  values  of  the  abscissas  and  ordinates  in  the  dia- 
gram by  any  number,  as  p.  Then  the  equation  of  the  parabola 
becomes 

PV  =  *?•  (2) 

,      Equation  (1)  may  be  written  in  the  form 

x*  +  (b  -p2)  x2  +p2x2  +  ex  +  d  =  0. 
Partly  substituting  py  for  x2, 

p2y2  +  (b  —  p^py  +p2x2  +  ex  +  d  =  0. 

*  For  the   following   exercises   a   graph   from  x  —  —  4   to  x  =  +  4   is 
sufficient. 


62 


GRAPHIC  ALGEBRA 


The  last  equation  is  easily  transformed  into  the  following 
one  (§  27)  : 


x  + 


+  [y  + 


2p 


0\  9 

-  jr  y  _ 


2^2 


+ 


5  —  p' 
2p 


2\  2 


-|.      (3) 


Xn  


Equation  (3)  represents  a  circle  whose  center  and  radius  are 
determined  by  the  formulae 

(4) 

(5) 
(6) 


2p2' 

_p2  —  6 
2p 

r2  =  x02  +  ?/02 


p2 


The  abscissas  of  the  points  of  in- 
tersection of  the  circle  (3)  and  the 
parabola  (2)  are  the  required  roots. 

Ex.    Solve 

xi  _  37  x2  -  24  x  4- 180  =  0. 

Since  obviously  x0  and  y0  are  very  large, 
multiply  the  values  on  the  two  axes  by  2 ; 
i.e.  make  p  =  2.  (The  new  values  are  given 
in  parentheses.) 

Applying  formulae  (4),  (5),  and  (6),  we 

have  : 

x0=3, 

2/o  =  10J, 
r  =  8.3+.* 
Construct  the  circle  and  measure  the  abscissas  of  the  points  of  inter- 
section.    Hence 

x  =  -5,  -3,  2,  6. 


EXERCISE   20 

Solve  graphically: 

1.  x4  -  45  jc2  -  40  a;  +    84  =  0.        3.    xA - 23 a2 - 18 x  +   40  =  0. 

2.  ic* - 42 x2- 64^  +  105  =  0.        4.    z4-37ar-24x  +  180  =  0. 
*  To  compute  »•,  use  table  of  squares  and  square  roots  in  Appendix  III. 


BIQUADRATIC  EQUATIONS 


63 


5.  a**- 75 x2- 70 x  +  144  =  0.       8.   z4-58a2  +  441  =0. 

6.  a4- 63 cc2  +  50 x  + 336=0.       9.   a?* -49 a?  +  36 a?  +  252  =  0. 

7.  z4-55a;2-30£  +  504  =  0.     10.   a,-4-49ar  -36x-  +  252  =  0. 

73.  Complex  roots.  If  an  equation  has  two  real  and  two 
complex  roots,  the  roots  may  be  found  by  a  method  similar  to 
the  one  employed  for  quadratics  and  cubics  (§§41  and  59). 

Let  the  equation  x2+px-\-q  =  0  have  equal  roots,  and  d  be 
a  positive  quantity.  Consider  the  following  three  equations 
which  are  supposed  not  to  contain  x3  when  simplified : 

(x-a)(x-b)(x2+px  +  q  +  d)  =  0,  (1) 

(x-a)(x-b)(x2+px  +  q)=0,  (2) 

(a:  -a){x-  b)  (x2  +px  +  q-d)=Q.  (3) 

Then  the  roots  are  (§  58)  respectively  : 

P 


a,b,-T}±V-d 


a,  b, 


a 


P       P. 


V 


,b,  —r>±  Vd. 


I.e.  the  roots  of  equation  (1)  are  complex,  but  they  may  be 
found  by  solving  (3)  instead. 

The  circles  C,  C,  and  C",  which  represent  respectively  equa- 
tions (1),  (2),  and  (3),  are  connected 
by  simple  geometric  relations,  which 
make  it  possible  to  construct  the 
third  circle  (C")  when  the  first  one 
(O)  is  given. 

1.  The  three  circles  C,  O,  and  C", 
pass  through  two  points,  P  and  P',  in 
the  parabola,  the  abscissas  of  P  and 
P'  being  a  and  b. 

Obviously  a  and  b  are  roots  of  the  equations  (1),  (2),  and  (3). 


64 


GRAPHIC  ALGEBRA 


2.  TJie  three  centers  C,  C,  and  C",  lie  in  the  perpendicular 
bisector  of  PP. 

3.  The  second  center,  C,  bisects  the  line  joining  the  other  two, 
Cand  C",orCC'=C'C". 

If  the  ordinate  of  C"  is  m,  then  the  ordinate  of  C  is  m  —  - ,  for  the 
coefficient  of  x2  in  (1)  is  greater  by  d  than  the  coefficient  of  x2  in  (2) 
(§  70).    Similarly,  the  ordinate  of  Cn  equals  m  +  -. 

74.  Hence  if  circle  C  is  given  and  it  intersects  the  parabola 
in  P  and  P,  construct  AB,  the  perpendicular  bisector  of  PP, 

and  in  AB  determine  C,  the  center 
of  the  circle  that  passes  through  P 
and  P,  and  touches  the  parabola  in 
another  point,  E.  Produce  CC  by 
its  own  length  to  C",  and  from  C", 
with  a  radius  equal  to  C"P,  draw  a 
circle.  This  circle  intersects  the 
parabola  in  two  other  points,  Q  and 
Q'.  If  the  abscissas  of  Q  and  Q'  are 
m  +  n  and  m  —  n,  the  required  roots 

are  respectively  m  +  n  V  — 1  and  m  —  n  V  —  1. 

The  abscissa  of  E  is  always  equal  to  m,  and  the  difference  of 
the  abscissas  of  Q  and  E  (or  E  and  Q')  is  equal  to  n. 

A  convenient  method  for  constructing  the  circle  C,  which 
touches  the  parabola  and  passes  through  P  and  P,  is  the  fol- 
lowing : 

Let  the  perpendicular  bisector  of  PP'  meet  PP  in  A,  and 
the  ?/-axis  in  B.  Produce  AB  by  its  own  length  to  D,  and 
let  the  ordinate  through  D  meet  the  parabola  in  E.  Then  E 
is  the  point  of  contact,  and  the  perpendicular  bisector  of  DE 
meets  CD  in  the  required  point  C". 


Ex.   Solve  the  equation 

x*  —  x2  —  4  x  —  4 


0. 


BIQUADRATIC  EQUATIONS 


65 


According  to  §  70, 


x0  =  2,y0  =  l,r  =  3. 


The  circle  drawn  from  (2,  1),  or  C,  as  center  with  a  radius  equal  to  3 
intersects    the    parabola    in     only   two 
points,  P  and  P'.    Hence  there  are  only 
two  real  roots,  viz.  —  1  and  2. 

Draw  AB,  the  perpendicular  bisector 
of  PP',  and  let  it  meet  the  y-axis  in  B. 
Produce  AB  by  its  own  length  to  D, 
and  draw  the  ordinate  DE,  E  being  a 
point  in  the  parabola.  The  perpen- 
dicular bisector  of  DE  meets  AB  in  C, 
the  center  of  the  second  circle.  Produce 
CC  by  its  own  length  to  C"  and  from 
C"  as  a  center  draw  a  circle  through  P 
and  P'.     This  circle,  C",  meets  the  parabola  in  two  other  points,  Q  and  Q'. 

The  abscissa  of  E,  i.e.  —  |,  is  the  real  part,  and  the  difference  of  the 
abscissas  of  E  and  Q  (or  E  and  Q'),  i.e.  1.3,  is  the  imaginary  part  of  the 
required  roots. 

Hence  the  roots  are  : 


\ 

/\ 

A 

/p 

m 

D 

a_„// 

\" 

ry 

(/ 

/ 

n 

P 

^ 

X 

*     -e      *V   0       i       *     'i 

■J±1.3aA 


1,2, 


and  —  1. 


EXERCISE   21 


Find  the  real  and  complex  roots  of  the  following  equations : 

6.  x4  +  3  x2  +  6  x  -  10  =0. 

7.  «*-  11a2- 14a?  +  24=0. 


1.  ^-8^  +  8^  +  15  =  0. 

2.  a4  -  7  a2  +  12.x  +  18  =  0. 

3.  x4-4x2-12x-9  =  0. 

4.  x4-5ar-10x-6  =  0. 

5.  x4  -  5  x2  -  4  a?  +  12  =  0. 


8.  x*-  10.x2  +  20  a -16  =  0. 

9.  x4-5x-'  +  10.x  -6  =  0. 
10.    x4-8x2+    8x  +  15=0. 


75.*    Solution  of  cubic  equations  by  means    of   the  standard 
parabola. 

To  solve  the  cubic 

xs  +  bx*  +  ex  +  d  =  0  (1) 

by  means  of  a  parabola,  multiply  by  (x  —  b),  i.e.  introduce  the  new  root  b. 

x*  +  (c  -  62)x2  +  (d  -  bc)x  -bd  =  0. 


66 


GRAPHIC  ALGEBRA 


Hence,  applying  §  70,  we  have 


d 


\P 

/ 

, 

JO 

10 

lo 

1- 

11 

10- 

f-1 

M 

\ 

/ 

\ 

/ 

\ 

/ 

\ 

U 

// 

\ 

5 

1 

V 

\ 

4 

4 

V 

\ 

./ 

/ 

A 

-i    , 

/ 

X' 

V 

V 

X 

^-5-4-^-2-1^123' 

o2  -  c  +  1 


(2) 

(3) 


The  formula  for  the  radius  is  not  neces- 
sary, since  the  circumference  must  pass 
through  the  point  of  the  parabola  whose 
abscissa  is  b,  i.e.  (b,  ft2). 

Thus,  to  solve 

a?  +  3  x2  -  6  x  -  8  =  0,  (4) 

either  multiply  by  x  —  3,  obtaining  xi  —  15  x2 
+  10  x  +  24  =  0,  or  apply  directly  formulae 
(2)  and  (3). 

Hence  x0  =  —  5, 

2/o  =  8. 

From  (—5,  8)  as  a  center  construct  a 
circle  passing  through  A,  i.e.  the  point  in  the 
parabola  whose  abscissa  is  3. 

Hence  the  roots  of  (4)  are  —  4,  —  1,  2. 

[For  examples  see  Exercise  16.] 

76.  Power  of  a  point  with  respect  to  a  circle.  If  r  is  the 
radius  of  a  circle  C,  and  d  the  distance  of  a  point  P  from  its 
center,  d2  —  r2  is  called  the  power  of  the 
point  P  icith  respect  to  circle  G. 

If  P  lies  without  the  circle  the  power 
is  j^ositive  and  equal  to  the  square  of 
the  tangent  drawn  from  P  to  the  circle.* 

If  the  point  lies  within  the  circle,  as 
P',  the  power  is  negative.     If  a  chord 
DE  is  drawn  perpendicular  to  CP\  the  power  of  P'  is  equal  to 
-  (P'D)2. 

77.  Values  of  a  biquadratic  function.     To  solve 

xA  +  bx>  +  cx  +  d  =  0,  (1) 

we  substitute  y  =  x2,  (2) 

*  Schultze  and  Sevenoak's  Geometry,  §  311. 


BIQUADRATIC  EQUATIONS 


67 


and  obtain  the  equation  of  a  circle  (§  70), 

(x-x0)2  +  (y-y0)2-r2  =  0. 

If  any  point  P,  whose  coordi- 
nates are  x'  and  y\  is  joined  to 
(x0,  y0)  i.e.  C,  we  have  (Geometry, 
§  310) 

2/o)2- 


(3) 


pc-  =  (x^-xQy  +  <y 

Hence  PC2  -  r2 
=  (x>  -  xoy  +  (y'  -  y0)2  -  r2. 

I.e.  if  we  substitute  the  coor- 
dinates of  any  point  P  in  the 
left  member  of  (3),  this  member  becomes  equal  to  the  power 
of  P  with  reference  to  circle  (3). 

If  the  point  P  is  located  in  the  parabola,  then  y'  —  x'2  and 
the  left  member  of  (3)  becomes  equal  to  the  left  member  of  (1). 
Hence,  the  value  of  the  function  (1)  for  any  particular  value  x'  is 
equal  to  the  poiver  of  point  (x',  x'2)  ivith  respect  to  circle  (3). 

78.  Thus,  to  find  the  various  values 
of  the  function 

y  =  x*  —  11  x2  —  4  x  +  6,    construct    a 
circle  so  that 

3^  =  2, 2fo  =  6,r  =  V34.    (§70) 

To  find  y  if  x  =  —  3|,  locate  in  the 
parabola  a  point  R,  whose  abscissa  is 
—  31,  and  draw  the  tangent  EH  to 
circle  C.  The  required  value  equals 
(RHy  =  (6-)2  =  36-. 

Similarly,  if  x  =  — 1-i,  locate  in  the 
parabola  a  point  S  whose  abscissa 
equals  — 1\,  and  draw  ST1.  CS,  then 
y  =  _  (ST)2.  To  find  (ST)2  graphi- 
cally, make  OA  =  ST,  then  the  ordi- 
nate AB  =  (ST)2,  or  the  function  equals  —  AB  =  —  7.7. 


68 


GRAPHIC  ALGEBRA 


Many  other  problems  relating  to  the  value  of  the  functions 

may  be  solved  by  such  a  diagram. 
Thus,  to  find  which  value  of  x  be- 
tween x  =  —  4  and  x  =  0  produces  the 
smallest  value  of  y,  determine  in  the 
parabola  the  point  nearest  to  C  by 
drawing  an  arc  EFD  from  C,  touch- 
ing the  parabola  in  F.  The  abscissa 
of  F,  i.e.  —  2.2,  is  the  required  value. 
Similarly,  we  can  determine  the 
greatest  value  of  the  function,  the 
value  of  x,  if  the  function  is  given,  etc. 

79.   The  graph  of  a  biquadratic  func- 
tion in  rectangular  coordinates  can  be 

constructed  by  means  of  §  77.  Since 
y  =  PC2—^,  construct  first  the  curve 
whose  ordinates  are  PC2 ;  i.e.  make  AP'  =  CP2,  EB'  =  CB2, 
etc.  (Use  table  in  Appendix  III,  and  draw  new  ordinates  on 
smaller  scale.) 

Locate  in  this  manner  a  sufficient 
number  of  points,  P',  B',  F,  D,  etc., 
and  draw  the  curve  P'B'FD.  Make 
O'O"  =  r2  and  through  0"  draw  XX' 
±0'0".  Then  the  curve  P'B'CD  re- 
ferred to  0 "  as  origin  is  the  required 
graph. 

EXERCISE  22 

If  y  =  x*-  3  x2  +  4  x  +  3,  find  graphi- 
cally the  value  of  y, 

1.  If  a:  =  2.  3.   Ifz  =  4. 

2.  Ifa  =  3.  4.    Ifa;  =  3.2. 

5.   Ifx  =  2.7. 


A     E    O' 


6.    Construct  the  graph  of  y  in  rectangular  coordinates. 


BIQ  UA  BRA  TIC  EQ  UA  TIONS 


69 


80.    Complete   biquadratic   equations.     The  complete   biquad- 
ratic equation 

x*  +  ax*  -f  bx2  +  cx  +  d  =  0  (1) 

is  transformed  into  another  equation  without  the  cubic  term 
by  the  substitution 

x  —  z 


a 
4" 


a 


The   resulting  equation  can  be  solved,  and  by  subtracting 

~  from  the  answers  the  roots  of  (1)  are  obtained. 
4  w 

Ex.   Solve 

aj*  +  4sc3-5£c2-22a;-8=:0.     (1) 

Substituting  x=z  —  \=z  —  1,         (2) 

(3_l)4  +  4(2-l)3-6(s-l)2-22(S-l) 

-8  =  0. 

Simplifying, 

24_ii22_4.~  +  g=o.* 
•'•    2o  =  2,y0  =  6:  r  =  V§4. 
Drawing  tbe  circle    and    measuring    tbe 
abscissas  of  the  points  of  intersection,  we 
obtain 

s=-3,  -1,  .6,3.4. 

Hence,  from  (2), 

x=-4,  -2,  -.4,  2.4. 

81.    In  most  cases,  the  method  of  the 
preceding  exercise  is   the   best.     It  is   possible,  however,  to 
derive  general  formulae. 


4A— 

/ 

1 

/ 

i 

f? 

P' 

A" 

S? 

A' 

•'• 

1    ( 

W  i      i      i 

If 


xA  +  aa3  +  bx2  +  ex  +  d  =  0, 


let 


#  =  z+p,  where  p  =  — 


a 


<1) 
(2) 


*  Students  who  are  familiar  with  the  general  method  for  removing 
the  second  term  should  of  course  use  this  method.  See  Schultze's 
Advanced  Algebra,  §  566. 


70 


GRAPHIC  ALGEBRA 


Then  equation  (1)  becomes 
yA  +  (6  p2  +  3  ap  +  b)z2  +  (4p3  +  3  ap2  +  2bp  +  c)z 

+p4  +  op3  +  &p2  +  op  +  d  =  0. 


(3) 


a 


Considering  that  p=  —  j,  we  can  easily  obtain  the  following 


values : 


a 

—2  ap2  —  2  ftp  —  c 

*° 2 ' 


2/o : 


2-3ap-2  6 


r2  =  #0?  +  2/o2  -  (j?4  +  ap3  +  ftp2  +  <%>  +  d)* 

Constructing  the  circle  (z0,  y0,  r)  and  measuring  the  abscissas 

of  the  points  of  intersection,  produces 

the  roots  of  (3),  and  hence  those  of  (1). 

Thus,  in  the  preceding  equation, 

a!*  +  4^-5^  -22  x  —  8  =  0, 

we  obtain 

p  =  -l. 

-8-10  +  22 


i 

k 

1 

Q 

t 

t 

/ 

H 

I 

/ 

Id" 

T 

' 

i« 

1 

/ 

11 

n 

//< 

W 

\ 

/ 

n 

\ 

/ 

V 

I 

\ 

/ 

\\ 

/ 

- 

\ 

■) 

I 

\ 

V 

1 

\ 

!'■ 

> 

\ 

X' 

V 

St 

X 

-  -3  -2  -1   ( 

UMM 

=2. 


2/o : 


2  +  12  +  10 


=  6. 


r2=4+36-(l-4-5+22-8)=34. 
Ex.    Solve        4  x*  + 16  x3  -  31  x2  - 
139  a? -60  =  0. 

Dividing  by  4,     x4  +  4  x3  -  Y  a;2  -  ^  x 
-15  =  0. 

*  Students  familiar  with  Calculus  can,  by  means  of  Taylor's  series, 
obtain 


-  f'(P)    „   -2-/"(P) 

—  o      >  ^o  —  ,. 


r2  =  *o2+yo2-/(P)- 


BIQUADRATIC  EQUATIONS  71 


Hence  p  =  —  1, 

z0  =  5f> 

2/o  =  7f, 
r  =  8.8. 
The  construction  of  the  circle  produces  the  values  z  =  —  3,  —  1£,  ^,  4. 
Hence  a;  =  z  +p  =  s  —  1. 

Or  x  =  -  4,  -  2£,  -  i,  3. 

EXERCISE  23 
Solve  the  equations : 

1.  x4  +  4ar5-9ar2-16x  +  20  =  0. 

2.  z4-4ar5-17aj2  +  24a:  +  36  =  0. 

3.  z4  -  8  af3  +  z2  +  78  .r- 72  =  0. 

4.  xi  +  8x*  +  Uxi-8x-15  =  0. 

5.  a;4  +  4a;3-4ar2-16a;  =  0. 

6.  o;4-2a;3-16ar9  +  2a;  +  15  =  0. 

7.  a4  +  4x3-21ar-64a  +  80  =  0. 

8.  x4  +  Sx3-3x2-62x  +  56  =  0. 

9.  a;4 -10  a3 +35  a,- -50  a; +  24  =  0. 

10.  a;4  +  3  a3  -  8  x2  -  12  x  +  16  =  0. 

11.  a:4-4ars-5ar  +  22a,--8  =  0. 

82.*    Solution  of  biquadratics  by  means  of  the  hyperbola  y  =  -. 

x 

Let  us  first  consider  the  equation 

x4  +  ax3  +  bx2  +  ex  +1  =  0.  (1) 

Partly  replacing  x  by-> 

y 

x_.ax.J>.c.j  _  o, 

y-2  y2  yl  y 

Or  x2  +  ax  +  b  +  q/  +  y2  =  0. 

Applying  §  27, 

7.e.  the  required  roots  are  determined  by  the  points  of  intersection  of  the 
standard  curve  y  =  -  (2)  and  the  circle  (3). 


72 


GRAPHIC  ALGEBRA 


The  circle  is  determined  by  the  formula 


xo  —  ~  n  '  ^o  — 


_^.r2 


r2  =  x02  +  2/02  -  6. 


83.*   The  equation 

x4  +  ax3  +  6x2  +  ex  +  d  =  0 
may  be  solved  by  the  circle  : 

a 


•2  dP 


2/o=   --^,^2  =  V  +  2/o2--i 


2d*  d2 

i 
The  abscissas  of  the  points  of  intersection  multiplied  by  d¥  are  the 

required  roots. 


84.*   Solution  of  a  cubic  by  the  hyperbola  y  =  - 

To  solve  xs  +  6x2  +  ex  +  d  =  0, 


(1) 


+  1  =  0. 


multiply  by  x  +  - ,  i.e.  introduce  the  new  root 

d  d 

Hence,  by  §  82,  we  have  to  construct  the  circle  that  is  determined  by  the 
f  ormulge : 


Vo 


The  formula  for  the  radius  is  not  necessary,  since  the  circle  must  pass 

through  the  point  f  — ,  —  d  ]  • 

Ex.    Solve  a;3- ar- 4 a  +  4  =  0. 

^0  =    —  2   (  — "  1   +  ?)    =  8» 

2/0=-K      4-1)  =  -i. 
From  (x0,  y0)  as  center  draw  a  circle 
through  (—  J,  —4),  r.e.  A.     By  meas- 
uring the  abscissas  of  the  other  points 
of  intersection,  we  obtain 
x  =  -  2,  1,  2. 

Note.  The  preceding  construction 
can  be  used  advantageously  for  large 
roots,  since  the  ordinates  do  not  become  as  large  as  in  the  case  of  the 
cubic  parabola. 


APPENDIX 


I.   GRAPHIC  SOLUTION   OF   PROBLEMS 


85.  Problems  are  usually  solved  in  algebra  by  expressing 
the  conditions  of  the  problems  in  the  form  of  equations.  By 
using  the  graphic  method,  however,  many  problems  can  be 
solved  directly,  without  obtaining  equations. 

The  fact  that  the  graph  of  two  proportional  variables  is  a 
straight  line  is  often  useful.  Thus,  if  x  and  y  are  the  coordi- 
nates of  a  point,  the  following  variables  are  represented  by 
straight  lines  :  x  —  time,  y  =  distance  covered  by  body  moving 
uniformly ;  x  =  time,  y  =  work  done  by  a  person ;  x  =  volume, 
y  =  weight  of  a  body ;  x  =  time,  y  =  quantity  of  water  flowing 
through  a  pipe  at  a  uniform  rate,  etc. 

86.  Uniform  motion.  To  represent  graphically  the  motion 
of  a  person  traveling  three 
miles  per  hour,  it  is  only 
necessary  to  locate  one 
point,  e.g.  (1,  3)  or  A,  and 
to  connect  this  point  to 
the  origin. 

The  increase  of  the  or- 
dinate per  hour  equals  the 
rate  of  travel,  i.e.  3  miles 
per  hour. 

Similarly,     CD    repre- 
sents the  motion  of  another  person  who  started  two  hours  later 
and  traveled  1\  miles  per  hour. 

EFOH  represents  graphically  that  a  third  person  had  a 
start  of  4  miles,  traveled  for  2  hours  at  the  rate  of  4  miles  per 

73 


F 

(0 

HI 

£ 

< 

F 

T 

0 

\ 

IU 

/ 

•J 

? 

o/ 

n 

— & 

I 

4    A 

H 

c 

HC 

URS 

X 

< 

) 

i 

M^ 

L       5         < 

r     6" 
l 

74 


APPENDIX 


hour,  then  rested  2  hours,  and  finally  returned  to  the  starting 
point  at  the  rate  of  2  miles  per  hour. 

IK  represents  graphically  the  motion  of  a  fourth  person 
who  started  3  hours  after  the  first  and  traveled  in  the  opposite 
direction  at  the  rate  of  1  mile  per  hour. 

Ex.  1.    A  and  B   start  walking   from  two  towns  15  miles 

apart,  and  walk  toward  each  other. 
A  walks  at  the  rate  of  3  miles  per 
hour,  but  rests  1  hour  on  the  way; 
B  travels  at  the  rate  of  4  miles  per 
hour  and  rests  3  hours.  In  how 
many  hours  do  they  meet  ? 

Construct  the  graphs  OA'A"A'"  and 
BB'B"B'".  The  abscissa  of  C,  the  point 
of  intersection,  is  the  required  time. 

Hence  A  and  B  meet  in  4^+  hours. 

Ex.  2.  A  stone  is  dropped  into  a 
well,  and  the  sound  of  its  impact 
upon  the  water  is  heard  at  the  top  of  the  well  5  seconds 
later.  If  the  velocity  of  sound  is  assumed  as  360  meters 
per    second,    and    g  =  10    meters,    how    deep    is    the    well  ? 

(A  body  falls  in  t  seconds  f- 12  meters.) 


B 

\ 

\ 

A/ 

12  \ 

B' 

B"J 

>B" 

4      V 

1 

/* 

\ 

7 

1 

. 

2       3       4  hou 
1        1 

RS 

Construct  the  graph  ODA  of  the  falling  body,  making  the  distances 
negative,  to  indicate  the  downward  mo- 
tion. Since  the  motion  of  the  sound  is  an 
upward  motion,  its  graph  CB  is  obtained 
by  joining  (4,-360)  and  (5,0).  The 
ordinate  of  the  point  of  intersection  D  is 
the  required  number. 

Hence  depth  of  well  =  110  meters. 

87.     Problems    relating   to    work 
done,  and  to  quantity  of  water  flowing  through  a  pipe,  are 
quite  similar  to  those  of  the  preceding  paragraph. 


APPENDIX 


75 


-;2 — 

n 

* 

s 

/) 
•   \ 

i. 

A 

B 

o 

z 

1- 
o 
1-  ^ 

c 

DAYS 

2       k       4       &       6  " 

Ex.  3.   A  can  do  a  piece  of  work  in  3  days,  and  B  in  6  days. 
In  how  many  days  can  both  do  it, 
working  together  ? 

Make  the  hour  equal  to  the  unit  of 
abscissas,  and  the  work  to  be  done  equal 
to  the  unit  of  the  ordinates.  Then  OA 
and  OB  represent  the  work  done  by  A 
and  B  respectively.  To  obtain  the 
graph  of  the  work  done  by  both  to- 
gether, we  add  the  ordinates  correspond- 
ing to  any  particular  time,  e.g.  3  hours ; 
i.e.  produce  CA  to  D  so  that  AD  =  CE. 
Then  OD  is  the  graph  of  the  work  done 
by  both,  and  the  required  time  is  equal  to  abscissa  of  S,  or  two  days. 

Hence  both  working  together  will  do  the  work  in  two  days. 

Ex.  4.  At  what  time  between  5  and  6  o'clock  are  the  hands 

of  a  clock  at  right  angles  ? 

Let  the  abscissa  represent  the  time 
from  5  to  6,  and  the  ordinate  the  hour 
spaces. 

It  can  easily  be  seen  that  AB  repre- 
sents the  motion  of  the  hour  hand.  A 
point  90°  distant  from  the  hour  hand 
moves  in  the  same  time  from  2  to  3  or 
from  8  to  9.  Hence  the  motion  of  such 
a  point  is  represented  by  CD  or  EF. 
But  the  graph  of  the  motion  of  the 
minute  hand    is    OG-.     Therefore    the 

abscissas  of  the  points  H  and  I  represent  the  required  time.     Or  the 

hands  are  at  right  angles  at  5  :  11  and  5  :  43|. 


EXERCISE  24 


1.  A  sets  out  walking  at  the  rate  of  3  miles  per  hour, 
and  3  hours  later  B  follows  on  horseback,  traveling  at  the 
rate  of  6  miles  per  hour.  After  how  many  hours  will  B  over- 
take A,  and  how  far  will  each  then  have  traveled  ? 


76 


APPENDIX 


2.  A  and  B  set  out  walking  at  the  same  time  in  the  same 
direction,  but  A  has  a  start  of  3  miles.  If  A  walks  at  the  rate 
of  2\  miles  per  hour,  and  B  at  the  rate  of  3  miles  per  hour, 
how  far  must  B  walk  before  he  overtakes  A  ?     ,     , 

h 

3.  A  train  traveling  30  miles  per  hour  starts  \]  of  an  hour 
before  a  second  train  that  travels  >'?■  miles  an  hour.  In  how 
many  hours  will  the  first  train  be  overtaken  by  the  second  ? 

4.  A  sets  out  walking  at  the  rate  of  3  miles  per  hour,  and 
one  hour  later  B  starts  from  the  same  point,  traveling  by  coach 
in  the  opposite  direction  at  the  rate  of  6  miles  per  hour. 
After  how  many  hours  will  they  be  27  miles  apart  ? 

5.  A  and  B  start  walking  at  the  same  hour  from  two  towns 
17^  miles  apart,  and  walk  toward  each  other.  If  A  walks  at 
the  rate  of  3  miles  per  hour  and  B  at  the  rate  of  4  miles  per 
hour,  after  how  many  hours  do  they  meet,  and  how  many  miles 
does  A  walk  ? 

6.  An  accommodation  train  runs  according  to  the  following 
schedule : ' 


Station 

Distance  feom  A 

Arrives 

Leaves 

A 

0 

2 

B 

10 

2:20 

2:24 

C 

15 

2  :32 

2:35 

D 

25 

2:50 

2:55 

E 

40 

3:20 

3  :21 

F 

50 

3:40 

An  express  train  leaves  A  at  2:15  and  reaches  Fat  3:25. 
Where  does  it  overtake  the  accommodation  train,  if  we  assume 
that  both  trains  move  uniformly  ? 


(a) 

A 

in 

(P) 

A 

in 

(<0 

A 

in 

(d) 

A 

in 

in 

6. 

in 

4. 

in 

6 

in 

5. 

APPENDIX  77 

7.  In  how  many  days  can  A  and  B  working  together  do  a 
piece  of  work  if  each  alone  can  do  it  in  the  following  number 
of  days  ? 

6,  B 

12,  B 

12,  B 

20,  B 

8.  A  can  do  a  piece  of  work  in  18  days,  B  in  9,  and  C  in  12 
days.  In  how  many  days  can  all  three  do  it  working  to- 
gether ? 

9.  At  what  time  between  2  and  3  o'clock  are  the  hands  of 
the  clock  together  ? 

10.  At  what  time  between  3  and  4  o'clock  are  the  hands  of 
a  clock  in  a  straight  line  and  opposite  ? 

11.  At  what  time  between  6  and  7  o'clock  are  the  hands  at 
right  angles  ? 

12.  A  cistern  can  be  filled  by  two  pipes  in  3  and  6  hours 
respectively.  In  how  many  hours  can  it  be  filled  by  the  two 
running  together  ? 

13.  A  cistern  can  be  filled  by  pipes  in  3,  4,  and  5  hours  re- 
spectively. In  how  many  hours  can  it  be  filled  by  all  three 
together  ? 

14.  A  stone  is  dropped  into  a  well  and  the  sound  of  its 
impact  upon  the  water  is  heard  at  the  top  of  the  well  (a)  4 
(6)  6  seconds  later.  If  the  velocity  of  sound  is  assumed  as  360 
meters  per  second,  and  g  =  10  meters,  how  deep  is  the  well  ? 

(A  body  falls  in  t  seconds  1 12  meters.) 


78 


APPENDIX 


II.     STATISTICAL   DATA   SUITABLE  EOR   GRAPHIC 

REPRESENTATION 

1.    Table  of  Population  (in  Millions)  of  United  States,  France, 
Germany,  and  British  Isles 


Tear 

U.S. 

France 

Germany 

British 

Isles 

1800 

5.3 

27.2 

22.0 

16.0 

1810     . 

7.2 

28.8 

23.4 

17.6 

1820     . 

9.6 

30.5 

26.2 

20.5 

1830     . 

12.9 

32.4 

29.7 

24.0 

1840     . 

17.0 

34.0 

32.4 

26.4 

1850     . 

23.2 

35.6 

35.2 

27.2 

1860     . 

31.4 

37.3 

38.1 

28.7 

1870     . 

38.6 

36.1 

40.5 

31.2 

1880     . 

50.2 

37.6 

45.2 

34.5 

1890     . 

62.6 

38.6 

49.4 

37.5 

1900     . 

76.3 

38.9 

56.4 

41.2 

2.    Arrival  of  Immigrants  (in  Ten  Thousands),  1891-1905 


From 

'91 
11 

'92 
13 

'93 
10 

'94 
6 

'95 

4 

'96 
3 

'97 
2 

'98 
2 

'99 
2 

'00 

'01 

'02 
3 

'03 

'04 

'05 

Germany 

2 

2 

4 

5 

4 

Italy 

8 

6 

7 

4 

4 

7 

6 

6 

8 

10 

14 

18 

23 

19 

22 

Russia 

5 

8 

4 

4 

3 

5 

3 

3 

6 

9 

9 

10 

14 

15 

18 

3.   Population  of  New  York  City 


1653 

1661 

1673 

1696 

1731 

1750 10,000 


1,120 

1756      .     .     . 

.     .     .     .       10,530 

1,743 

1771      .     .     . 

.     .     .     .       21,865 

2,500 

1774      .     .     . 

.     .     .     .       22,861 

4,455 

1786      .     .     . 

.     .     .     .      23,688 

8,256 

1790      .     .     . 

.     .     .     .       33,131 

0,000 

1800      .     .     . 

.     .     .     .       60,489 

APPENDIX 


79 


3.  Population  of  New  York  City  —  Continued 


1805 75,587 

1810 96,373 

1816 100,619 

1820 123,706 

1825 166,136 

1830 202,589 

1835 253,028 

1840 312,710 

1845 358,310 

1850 515,547 

1855 629,904 

1860 813,669 

1865 726,836 

1870 942,292 


1875 1,041,886 

1880 1,206,299 

1890 1,515,301 

1893 1,891,306 

1898  (all  Boro's)       .     .  3,350,000 

1899  (all  Boro's)       .     .  3,549,558 

1900  (all  Boro's)       .     .  3,595,936 

1901  (all  Boro's)       .     .  3,437,202 

1902  (all  Boro's)       .     .  3,582,930 

1903  (all  Boro's)       .     .  3,632,501 

1904  (all  Boro's)       .     .  3,750,000 

1905  (all  Boro's)       .     .  3,850,000 

1906  (all  Boro's)       .     .  4,014,304 


Population    (in   Hundred   Thousands)    of    Illinois,    Massachu- 
setts, New  York,  and  Virginia 


State 

1800 

1810 

1820 

18.30 

1840 

1850 

8.5 

1860 
17.1 

1870 

1880 
30.8 

1890 

1900 

Illinois 

.5 

1.6 

4.8 

25.3 

38.3 

48.2 

Mass. 

3.4 

3.8 

4.0 

4.5 

4.7 

5.8 

6.9 

7.8 

9.3 

10.4 

11.9 

N.  York 

6.9 

9.6 

13.7 

19.2 

24.3 

31.0 

38.8 

43.8 

50.8 

60.0 

72.7 

Virginia 

8.8 

9.7 

10.7 

12.1 

12.4 

14.2 

16.0 

12.3 

15.1 

16.6 

18.5 

5.   Table  of  Mortality 


Com- 

Number 

Deaths 

Number  of 

Number 
Dying  an- 
nually out  of 
Each  1000 

pleted 
Age 

Surviving  at 
Each  Age 

in  Each 
Year 

Years 
Expectation 

10 

100,000 

749 

48.7 

7.49 

11 

99,251 

746 

48.1 

7.52 

12 

98,505 

743 

47.4 

7.54 

13 

97,762 

740 

46.8 

7.57 

14 

97,022 

737 

46.2 

7.60 

15 

96,285 

735 

45.5 

7.63 

16 

95,550 

732 

44.9 

7.66 

17 

94,818 

729 

44.2 

7.69 

18 

94,089 

727 

43.5 

7.73 

19 

93,362 

725 

42.9 

7.77 

20 

92,637 

723 

42.2 

7.81 

80 


APPENDIX 


5.  Table  of  Mortality — Continued 


Com- 

Number 

Deaths 

Number  of 

Number 
Dying  an- 
nually out  of 
Each  1000 

pleted 
Age 

Surviving  at 
Each  Age 

in  Each 
Tear 

Tears  Ex- 
pectation 

21 

91,914 

722 

41.5 

7.86 

22 

91,192 

721 

40.9 

7.91 

23 

90,471 

720 

40.2 

7.96 

24 

89,751 

719 

39.5 

8.01 

25 

89,032 

718 

38.8 

8.07 

26 

88,314 

718 

38.1 

8.13 

27 

87,596 

718 

37.4 

8.20 

28 

86,878 

718 

36.7 

8.26 

29 

86,160 

719 

36.0 

8.35 

30 

85,441 

720 

35.3 

8.43 

31 

84,721 

721 

34.6 

8.51 

32 

84,000 

723 

33.9 

8.61 

33 

83,277 

726 

33.2 

8.72 

34 

82,551 

729 

32.5 

8.83 

35 

81,822 

732 

31.8 

8.95 

36 

81,090 

737 

31.1 

9.09 

37 

80,353 

747 

30.4 

9.23 

38 

79,611 

749 

29.6 

9.41 

39 

78,862 

756 

28.9 

9.59 

40 

78,106 

765 

28.2 

9.79 

41 

77,341 

774 

27.5 

10.01 

42 

76,567 

785 

26.7 

10.25 

43 

75,782 

797 

26.0 

10.52 

44 

74,985 

812 

25.3 

10.83 

45 

74,173 

828 

24.5 

11.16 

46 

73,345 

848 

23.8 

11.56 

47 

72,497 

870 

23.1 

12.00 

48 

71,627 

896 

22.4 

12.51 

49 

70,731 

927 

21.6 

13.11 

50 

69,804 

962 

20.9 

13.78 

51 

68,842 

1,001 

20.2 

14.54 

52 

67,841 

1,044 

19.5 

15.39 

53 

66,797 

1,091 

18.8 

16.33 

54 

65,706 

1,143 

18.1 

17.40 

55 

64,563 

1,199 

17.4 

18.57 

56 

63,364 

1,260 

16.7 

19.89 

57 

62,104 

1,325 

16.1 

21.34 

58 

60,779 

1,394 

15.4 

22.94 

59 

59,385 

1,468 

14.7 

24.72 

60 

57,917 

1,546 

14.1 

26.69 

61 

56,371 

1,628 

13.5 

28.88 

62 

54,743 

1,713 

12.9 

31.29 

63 

53,030 

1,800 

12.3 

33.94 

64 

51,230 

1,889 

11.7 

36.87 

APPENDIX 


81 


6.  Table  of  Mortality  —  Continued 


Com- 

Number 

Deaths 

Number  of 

Number 
Dying  an- 
nually out  of 
Each  1000 

pleted 

Surviving  at 

in  Each 

Years  Ex- 

Age 

Each  Age 

Year 

pectation 

65 

49,341 

1,980 

11.1 

40.13 

66 

47,361 

2,070 

10-5 

43.71 

67 

45,291 

2,158 

10.0 

47.65 

68 

43,133 

2,243 

9.5 

52.00 

69 

40,890 

2,321 

9.0 

56.76 

70 

38,569 

2,391 

8.5 

61.99 

71 

36,178 

2,448 

8.0 

67.67 

72 

33,730 

2,487 

7.6 

73.73 

73 

31,243 

2,505 

7.1 

80.18 

74 

28,738 

2,501 

6.7 

87.03 

75 

26,237 

2,476 

6.3 

94.37 

76 

23,761 

2,431 

5.9 

102.31 

77 

21,330 

2,369 

5.5 

111.06 

78 

18,961 

2,291 

5.1 

120.83 

79 

16,670 

2,196 

4.8 

131.73 

80 

14,474 

2,091 

4.4 

144.47 

81 

12,383 

1,964 

4.1 

158.61 

82 

10,419 

1,816 

3.7 

174.30 

83 

8,603 

1,648 

3.4 

191.56 

84 

6,955 

1,470 

3.1 

211.36 

85 

5,485 

1,202 

2.8 

235.55 

86 

4,193 

1,114 

2.5 

265.68 

87 

3,079 

933 

2.2 

303.02 

88 

2,146 

744 

1.9 

346.69 

89 

1,402 

555 

1.7 

395.86 

90 

847 

385 

1.4 

454.55 

91 

462 

246 

1.2 

532.47 

92 

216 

137 

1.0 

634.26 

93 

79 

58 

.8 

734.18 

94 

21 

18 

.6 

857.14 

95 

3 

3 

.5 

1,000.00 

6. 

Railway  Accidents  in  the 

United  States 

Total 

Killed 

Injured 

1897 

6,437 

36,731 

1898 

6,859 

40,882 

1899 

7,123 

44,620 

1900 

7,865 

50,320 

1901 

8,455 

53,339 

1902 

8,588 

64.662 

1903 

9,840 

76,553 

1904 

10,046 

84, 155 

82 


APPENDIX 


7.  Amount  of  $1  at  Compound  Interest  from  One  to  Thirtt  Years 


Tears 

3J  Per  Cent 

4  Per  Cent 

5  Per  Cent 

6  Per  Cent 

1 

1.035 

1.040 

1.050 

1.060 

2 

1.071 

1.081 

1.102 

1.123 

3 

1.108 

1.124 

1.157 

1.191 

4 

1.147 

1.169 

1.215 

1.262 

5 

1.187 

1.216 

1.276 

1.338 

6 

1.229 

1.265 

1.340 

1.418 

7 

1.272 

1.315 

1.407 

1.503 

8 

1.316 

1.368 

1.477 

1.593 

9 

1.362 

1.423 

1.551 

1.689 

10 

1.410 

1.480 

1.628 

1.790 

11 

1.460 

1.539 

1.710 

1.898 

12 

1.511 

1.601 

1.795 

2.012 

13 

1.564 

1.665 

1.885 

2.132 

14 

1.618 

1.731 

1.979 

2.260 

15 

1.675 

1.800 

2.078 

2.396 

16 

1.734 

1.873 

2.182 

2.540 

17 

1.794 

1.947 

2.292 

2.692 

18 

1.857 

2.025 

2.406 

2.854 

19 

1.922 

2.106 

2.527 

3.025 

20 

1.989 

2.191 

2.653 

3.207 

21 

2.059 

2.278 

2.786 

3.399 

22 

2.131 

2.369 

2.925 

3.603 

23 

2.206 

2.464 

3.071 

3.819 

24 

2.283 

2.563 

3.225 

4.048 

25 

2.363 

2.665 

3.386 

4.291 

26 

2.446 

2.772 

3.555 

4.549 

27 

2.531 

2.883 

3.733 

4.822 

28 

2.620 

2.998 

3.920 

5.111 

29 

2.711 

3.118 

4.116 

5.418 

30 

2.806 

3.243 

4.321 

5.743 

APPENDIX 


83 


8.  Amount  of  $1  Annually  Deposited  at  Compound  Interest 


Years 

3 J  Pbk  Cent 

4  Per  Cent 

5  Per  Cent 

6  Per  Cent 

1 

1.000 

1.000 

1.000 

1.000 

2 

2.035 

2.040 

2.050 

2.060 

3 

3.106 

3.121 

3.152 

3.183 

4 

4.215 

4.246 

4.310 

4.374 

5 

5.363 

5.416 

5.525 

5.637 

6 

6.550 

6.633 

6.801 

6.975 

7 

7.779 

7.898 

8.142 

8.393 

8 

9.052 

9.214 

9.549 

9.897 

9 

10.368 

10.582 

11.026 

11.491 

10 

11.731 

12.006 

12.577 

13.180 

11 

13.142 

13.486 

14.206 

14.971 

12 

14.602 

15.025 

15.917 

16.869 

13 

16.113 

16.626 

17.713 

18.882 

14 

17.677 

18.291 

19.598 

21.015 

15 

19.296 

20.023 

21.578 

23.276 

16 

20.971 

21.824 

23.657 

25.672 

17 

22.705 

23.697 

25.840 

28.212 

18 

24.500 

25.645 

28.132 

30.905 

19 

26.357 

27.671 

30.539 

33.760 

20 

28.280 

29.778 

33.066 

36.785 

21 

30.270 

31.969 

35.719 

39.992 

22 

32.328 

34.248 

38.505 

43.392 

23 

34.460 

36.617 

41.430 

46.995 

24 

36.666 

39.082 

44.502 

50.815 

25 

38.949 

41.645 

47.727 

54.864 

26 

41.313 

44.311 

51.113 

59.156 

27 

43.759 

47.084 

54.669 

63.705 

28 

46.290 

49.967 

58.402 

68.528 

29 

48.910 

52.966 

62.322 

73.639 

84 


APPENDIX 


III.    TABLES 


TABLE   1 


Squares,  Cubes,  Square  Roots  and  Reciprocals  of  Numbers 

from  1  to  100 

The     squares,    cubes,    and    reciprocals    of    decimal    frac- 
tions  can  be  obtained  by  shifting  the  decimal  point.     Thus 

4.22  =  17.64,  4.23  =  74.088,  —  =  .24.     For  square  roots,  how- 
ever, this  method  fails,  and  Table  2  has  to  be  used. 


■ 

X* 

as3 

•£ 

X 

xr 

I 

I 

I 

I. OOO 

I. OOO 

I 

2 

4 

8 

I.4I4 

.500 

2 

3 

9 

27 

1-732 

•333 

3 

4 

16 

64 

2.000 

.250 

4 

5 

25 

125 

2.236 

.200 

5 

6 

36 

216 

2.449 

.167 

6 

7 

49 

343 

2.646 

•  143 

7 

8 

64 

512 

2.828 

.125 

8 

9 

81 

729 

3.OOO 

.111 

9 

IO 

1  00 

1  000 

3.162 

.100 

10 

ii 

1  21 

I331 

3-3I7 

.091 

11 

12 

144 

1  728 

3-464 

.0S3 

12 

13 

1  69 

2197 

3.606 

.077 

13 

14 

1  96 

2744 

3-742 

.071 

14 

15 

2  25 

3  375 

3-873 

.067 

15 

16 

256 

4096 

4.000 

.063 

16 

17 

289 

49i3 

4-123 

•059 

17 

18 

324 

5832 

4-243 

.056 

18 

19 

361 

6859 

4-359 

•053 

19 

20 

400 

8  000 

4.472 

.050 

20 

X 

St* 

X3 

V* 

1 

X 

x 

APPENDIX 


85 


X 

X* 

Xs 

v'x; 

1 

x 

X 

21 

441 

9  261 

4-583 

.048 

21 

22 

484 

10648 

4.690 

•045 

22 

23 

529 

12  167 

4.796 

.043 

23 

24 

576 

I3824 

4.899 

.042 

24 

25 

625 

I5625 

5.000 

.040 

25 

26 

676 

I7576 

5-°99 

.039 

26 

27 

729 

19  683 

5-I96 

•037 

27 

28 

784 

21  952 

5.292 

.036 

28 

29 

841 

24  389 

5-385 

•034 

29 

30 

900 

27  OOO 

5-477 

•033 

SO 

31 

9  61 

2979I 

5.568 

.032 

31 

32 

10  24 

32768 

5-657 

.031 

32 

33 

1089 

35  937 

5-745 

.030 

33 

34 

n  56 

39  304 

5-83I 

.029 

34 

35 

1225 

42875 

5.916 

.029 

35 

36 

12  96 

46656 

6.000 

.028 

36 

37 

1369 

50653 

6.083 

.027 

37 

38 

1444 

54872 

6.164 

.026 

38 

39 

15  21 

59  319 

6.245 

.026 

39 

40 

1600 

64000 

6.325 

.025 

40 

4i 

1681 

68921 

6.403 

.024 

4i 

42 

1764 

74088 

6.481 

.024 

42 

43 

1849 

79  507 

6-557 

.023 

43 

44 

1936 

85  184 

6.633 

.023 

44 

45 

2025 

91  125 

6.708 

.022 

45 

46 

21  16  "• 

97  336 

6.782 

.022 

46 

47 

22  09 

103  823 

6.856 

.021 

47 

48 

2304 

1 10  592 

6.928 

.021 

48 

49 

24  01 

117  649 

7.000 

.020 

49 

5° 

2500 

125  000 

7.071 

.020 

50 

5i 

2601 

132  651 

7. 141 

.020 

51 

52 

2704 

140  608 

7. 211 

.019 

52 

53 

2809 

148877 

7.280 

.019 

53 

54 

29  16 

157464 

7-348 

.019 

54 

55 

3025 

166375 

7.416 

.018 

55 

56 

3136 

175  616 

7-483 

.018 

56 

57 

3249 

185  193 

7-55° 

.018 

57 

58 

3364 

195  112 

7.616 

.017 

58 

59 

34  8i 

205  379 

7.681 

.017 

59 

60 

3600 

216  000 

7.746 

.017 

60 

X 

as2 

SC3 

^x 

1 

X 

X 

86 


APPENDIX 


X 

X* 

X3 

*X 

1 

X 

X 

61 

37  21 

226981 

7.810 

.016 

61 

62 

38  44 

238  328 

7.874 

.016 

62 

63 

39  69 

250  047 

7-937 

.016 

63 

64 

4096 

262  144 

8.000 

.016 

64 

65 

4225 

274  625 

8.062 

.015 

65 

66 

43  56 

287  496 

8.124 

.015 

66 

67 

4489 

300  763 

8.185 

.015 

67 

68 

46  24 

3H432 

8.246 

.015 

68 

69 

4761 

338  5°9 

8.307 

.014 

69 

70 

4900 

343  000 

8.367 

.014 

70 

7i 

50  4I 

3579" 

8.426 

.014 

7i 

72 

5184 

373  248 

8.485 

.014 

72 

73 

53  29 

389017 

8-544 

.014 

73 

74 

54  76 

405  224 

S.602- 

.014 

74 

75 

5625 

421  875 

8.660 

.013 

75 

76 

57  76 

438  976 

8.718 

.013 

76 

77 

59  29 

456  533 

8-775 

.013 

77 

78 

6084 

474  552 

8.832 

.013 

78 

79 

6241 

493  039 

8.888 

.013 

79 

80 

6400 

5 1 2  000 

8-944 

.013 

80 

81 

6561 

53i  441 

9.000 

.012 

81 

82 

6724 

551368 

9-055 

.012 

82 

83 

6889 

57i  787 

9.110 

.012 

83 

84 

7056 

592  704 

9.165 

.012 

84 

85 

7225 

614  125 

9.219 

.012 

85 

86 

73  96 

636  056 

9.274 

.012 

86 

87 

7569 

658  5°3 

9-327 

.Oil 

87 

88 

77  44 

681  472 

9.381 

.Oil 

88 

89 

79  21 

704  969 

9-434 

.Oil 

89 

90 

81  00 

729000 

9.487 

.Oil 

90 

9i 

8281 

753  571 

9-539 

.Oil 

9i 

92 

8464 

778  688 

9-592 

.Oil 

92 

93 

8649 

804  357 

9.644 

.Oil 

93 

94 

8836 

830  584 

9.695 

•Oil 

94 

95 

9025 

857  375 

9-747 

.Oil 

95 

96 

92  16 

884  736 

9.798 

.010 

96 

97 

9409 

912673 

9.849 

.010 

97 

98 

9604 

941  192 

9.899 

.010 

98 

99 

^801 

970  299 

9-95° 

.010 

99 

100 

100  00 

1  000  000 

10.000 

.010 

100 

* 

x2 

X* 

Vx 

I 

X 

X 

APPENDIX 


87 


TABLE   2 


Square  Roots  of  Numbers  from  1  to  9.9 


o.o 

O.I 

0.2 

0.3 

0.4 

0-5 

•0.6 

0.7 

0.8 

0.9 

o 

o.ooo 

0.316 

0.447 

0.548 

0.632 

0.707 

°-775 

0.837 

0.894 

0.949 

I 

I.OOO 

1.049 

1.095 

1. 140 

1. 183 

1.225 

1.265 

1.304 

1.342 

i-37* 

2 

1.414 

1.449 

1.483 

1-517 

1-549 

1.581 

1.612 

1.643 

I.673 

1-703 

3 

!-732 

1. 761 

1.789 

1.817 

1.844 

1.871 

1.897 

1.924 

1.949 

J-975 

4 

2.000 

2.025 

2.049 

2.074 

2.098 

2.121 

2.145 

2.168 

2.191 

2.214 

5 

2.236 

2.258 

2.280 

2.302 

2.324 

2-345 

2.366 

2.387 

2.408 

2.429 

6 

2.449 

2.470 

2.490 

2.510 

2.530 

2-55° 

2.569 

2.588 

2.608 

2.627 

7 

2.646 

2.665 

2.683 

2.702 

2.720 

2-739 

2-757 

2.775 

2-793 

2.81 1 

8 

2.828 

2.846 

2.864 

2.881 

2.898 

2.915 

2-933 

2.950 

2.966 

2.983 

9 

3.000 

3017 

3-033 

3.050 

3.066 

3.082 

3.098 

3-"4 

3-130 

3.146 

ANSWERS  TO  EXERCISES 


Exercise  1.    Pages  2,  3 

5.  5.66.  6.    5.  7.   In  a  line  II XX'  passing  through  (0,  4). 

8.  In  the  ?/-axis.  9.   In  the  x-axis. 

10.  The  line  II XX'  passing  through  (0,  3). 

11.  The  ordinate.  12.    (0,  0). 

Exercise  2.    Pages  6,  7,  8 

1.  (a)  6°,  5.9°,  5.25°,  2.5°.  (6)  1  :  40  p.m.  and  5  p.m;  1  p.m.  and 

6  p.m.  ;  11  a.m.  and  8.40  p.m.  ;  10  p.m.  ;  9.20  p.m. 

(c)  3.15  p.m.  (d)  7+°.  (e)  1  p.m.  to  6  p.m. 
(/)  12  m.  to  12.30  p.m.  ;  6.40  p.m.  to  7.20  p.m. 

(g)  11  a.m.  to  9.20  p.m.     (h)  9.20  p.m.  on.   (£)  4.75°.    (&)  6  p.m. 

(Z)  11  a.m.  to  3.15  p.m.  (m)  3.15  p.m.  on. 

(n)  Between  3  p.m.  and  4  p.m. 

(o)  Between  12  m.  and  1  p.m.  ;  and  between  11  a.m.  and  12  m. 

2.  (a)  San  Francisco.        (&)  Bismarck.        (c)  April  20  and  Sept.  20. 

(d)  During  April.         (e)  25°. 

3.  (c)  18° C.  (d)  8.1  grams.       (e)  15  grams. 


Exercise  3.    Pages  10,  11 

(rt)  12.25.                   (6)  2.25.                   (c)  7.84. 

(d)  3.61, 

(e)  2.5.                       (/)  3.5.                   (g)  2.24. 

(ft)  .55. 

(a)  4.25,  -  1.75,  -  1.75.     (6)  2  ;  3.73  and  .27  ; 

3.87  and  .13. 

(c)   -2.            (<f)  2.            (e)  3.41  and  .59. 

(/)  3.41  and  .59. 

(g)  3  and  1.                           (h)  0  and  4. 

(a)  2.75,  -3.25,  1.5.            (6)  3.24,  -1.24. 

(c)  3.          (d)  1 

(e)  2.73,  -  .73.      (/)  2.73,  -  .73.      (g)  2.4,  - 

.4.      (ft)  2.4,  -A 

25. 
26. 

27. 

28.  (6)  31.25  meters.         (c)  2.24  seconds. 

Exercise  4.    Page  12 

7.  (&)  -18i°C,  -12|°C,  -10°C,  0°C.     (c)  14°F.,  32°F.,  33|°F. 

89 


Exercise  5.    Pages  15,  16 

1. 

1.75. 

5.   3,  -  2. 

9.  .7,  -5.7. 

13. 

-1.93,  2.93. 

2. 

-2.5. 

6.   2.79,  - 

■1.79.         10.  5.54,  -.54. 

14. 

-1.92,3.92. 

3. 

6. 

7.   3.83,  - 

-1.83.         11.  4.37,  -1.37. 

15. 

-5.62,  .62. 

4. 

2.67. 

8.  3,  3. 

12.  -2.16,  4.16. 

16. 

-1.31,  3.31. 

17. 

-1.53, 

-  .35, 1.88. 

20.  1.21,  2  imag.              23. 

-3.1, 

3.5,  4.6. 

18. 

-4.05, 

2  imag. 

21.  1.78,  2  imag.              24. 

-.39, 

5.44,  7.95. 

19. 

-2.11, 

.25,  1.86. 

22.  -  1.94,  .55,  1.39.      25. 

±.94, 

±3.02. 

26.    -2.5, 1.73,  2  imag.  28.    -  2.99,  -  1.15,  .21,  1.9,  3.05. 

27. -.97,  .85,  2.15,  3.97.  29.    1.38. 

30.  (a)    -4.51,-1.75,1.26.  (6)    -4.12,-2.4,1.52. 
(c)    -  4.78,  -  1.14,  .92.  (d)    -  5.19,  2  imag. 

(e)  3.     (/)  -10  to  8.5+.     (g)  -  10  or  8.5+.     (h)  8.5.     (i)   -3.33. 

31.  (a)    -  2.84,  .44,  2.4.  (b)    -  2.65,  0,  2.65. 

(c)    -  3,  1,  2.  (d)    -  1.68,  -  1.38,  3.05. 

(e)    -  3.49,  2  imag.  (/)    1,  3,  3,  1.            (g)   2,  2,  0. 

Exercise  7.    Page  22 

1.  x  =  2.8,  y  =  —  .1.         8.  Parallel.                       11.  x  =  4,  y  =  3. 

2.  x  =  2,  y  =  1.                 9.  x  -  3.6,  y  =  —  1.6.            x  =  —  3,  y  =  -  4. 

3.  x  =  -  .8,  y  =  —  2.8.  x  =  -  1.6,  2/  =  3.6.    12.  x  =  4,  j/  =  2. 

4.  x  =  2.6,  y  =  .7.            10.  x  =  3,  ?/  =  2.                        x  =  —  2,  y  =  —  4. 

5.  x  =  1.5,  y  =  .5.  x  =  2,  y  =  3. 

13.  x  =  4.3,  J/  =  1.4.  14.  x  =  2.  3,  y  =  1.15. 

x  =  -  1.8,  y  =  -  3.4.  x  =  -2.3,  y=  -1.15. 

15.  x  =  ±  4.8,  ?/  =  ±  1.3.  16.  x=  ± 3,  y  =  ±  1. 

x  =  ±  1.3,  y  =  ±  4.8.  x=±00,2/=:Foo- 

Exercise  8.    Page  24 

1.  x  =  1.82,  y  =  -  .82.  6.  x  =  2, 2/  =  4. 
x  =  -  .82,  2/  =  1.82.  x  =  0,  2/  =  0. 

2.  x  =  4,  y  =  0.  T.  x  =  3,  y  =  0. 

x  =  0,  2/  =  —  4.  x  =  1,  2/  =  —  2. 

8.  as  =  —  7, 2/  =  —  1.  8.  x  =  1,  2/  =  ±  3. 

x  =  1,  2/  =  7.  9.  x  =  .22,  y  =  1.72. 

4.  x  =  2.96,  2/  =  .48.  x  =  -1.72,  2/  =  -  .22. 
X  =  -  2.16,  2/  =  -  2.08.  10.  x  =  0,  2/  =  ±  1. 

5.  x  =  -  3.4,  2/  =  -  2.1. 
x  =  .03,  y  =  3.95. 


ANSWERS 


91 


Exercise  9.    Page   28 


.i-"5 


1. 

3,  -  2.          5, 

4,-2. 

( 

3.    -  6,  -  .67. 

12.   2.7,   -5.1. 

2. 

1,-2.           6. 

1.24,  -  3.24. 

10.    1.8,  -  2.8. 

13.    2.1,  -4.6. 

3. 

6,  -  3.           7. 

7.1,  -2.1. 

11.    4.2,  -2.2. 

14.    1.5,  -  .7. 

4. 

2,  -  5.           8. 

.95,  6.3. 

Exercise  10. 

Pages  31,  32 

1. 

-  60,  75. 

7.    -  20,  30. 

13.    -35,6. 

19.    -.2,  .3. 

2. 

-  50,  60. 

8.    -  16,  8. 

14.    -  72.2,  5.5. 

20.    -  .6,  .2. 

3. 

-  60,  -  20. 

9.    -  30,  60. 

15.    -.225,  .525.      21.     -  .2,  .1. 

4. 

-  60,  20. 

10.    -  55,  22. 

16.    -  .136,  .736.      22.     -  .25,  .6. 

5. 

-  80,  60. 

11.    -  28.3,  26.8 

17.    -1.425,  .17 

5. 

6. 

-  70,  -  10. 

12.    -  33.33,  30. 

18.    -  .311,  .161. 

Exercise  11. 

Pages  34,  35 

1. 

13,6,  -2.75, 

-  2.75. 

7. 

—  1  when  x  = 

2. 

2. 

-  2.25,  -  3,  - 

-3,3. 

8. 

—  30  when  x  = 

=  -5. 

3. 

2.91,  -2.25, 

-  17.25. 

9 

—  15.25  when 

x  =  -3.5. 

4. 

31.25,  130.96, 

47.24. 

10 

,    —  4.25  when  x 

;  =  2.5. 

5. 

4.87,  -2.87. 

11. 

7,  130.5,  147. 

6. 

-  9.47,  -  .53. 

Exercise  12. 

Page  39 

1.   5,  5. 

5.  4±3z. 

9. 

-1.5  ±4.97  i. 

13.    -.5±1.12i. 

2.   5  ±  2  i. 

6.   5  ±  2  i. 

10. 

-  4.5  ±  3.97  tf. 

14.   1.5  ±  2  i. 

3.    -  2  ±  2  i. 

7.    _3.5±2.96i. 

11. 

-.5 ±.87  i. 

4.   -  4  ±  2  i. 

8.   2.5  ±  2.96  i. 

12. 

-1,-1. 

Exercise  13. 

Page  41 

1.   5,  -3. 

3.    5,  1. 

5.    1  ±  3  i. 

2.    3,  -2. 

4.    -4,  -1. 

6.    -1.27,  -4.73. 

Exercise  14. 

Page  45 

1.  2. 

9.    2.7. 

17. 

1.1. 

2.  3. 

10.   6.6. 

18. 

8. 

3.    -2. 

11.    -2.6. 

19. 

4.5,4.5,  -9. 

4.    -3. 

12.   3.3. 

20. 

-11. 

5.    1,2,  -3. 

13.   4.5 

21. 

-  5,  -  5,  10. 

6.    -21,  -1, 

3£.          14.    -4.6. 

22. 

-11.9. 

7.    -3.3. 

15.    -4.5 

23. 

4.1. 

8.    3.2. 

16.    -.6, 

6.7, 

-  5.2.            24. 

-  16.5. 

92 


l. 

2. 
3. 
4. 
5. 
6. 
7. 

1. 
2. 
3. 
4. 
5. 
6. 


^sio^s 

Exercise  15. 

Page  48 

-55,  -41.6,  -33.7,  40.9. 

8.    3.4. 

-18.5,  -17.4,8,  1.4. 

9.    -16. 

-24,  -11,  -20.6,  6.9. 

10.    14. 

-  454,  -  82,  30,  326. 

13.   101,  73.7,  -.2. 

3.2. 

14.    -133,  -445,  - 

-67. 

2.4,  .4,  -2.1. 

15.    -31. 

-3.6. 

Exercise  16.    Page  52 


-1,-1,  2. 
-2,1,1. 

-  2, 1  ±  2  i. 

2,  -1  ±  .5  i. 
2,  -1-1-1.5*. 

-  2,  1  ±  3.46  i. 


7.  -4,  2  ±1.73*. 

8.  _  7,  3.5  ±  5.27  i. 

9.  -  1.5,  .75  ±  .43  i. 

10.  2,  -  1  ±  .87  i. 

11.  1,  -  .5  ±2.18  t. 

12.  -  1.46,  .73  ±  1.9  i. 


Exercise  17.    Pages  56,  57 


1.  -1,1,3. 

2.  1,  3,  5. 

3.  -  1,  2,  5. 

4.  1,  2,  5. 

5.  -  1,  -  2,  -  4. 

6.  -2,  1,  3. 

7.  -  1.52,  .43,  3.09. 

8.  -1.18,  2  imag. 

9.  -4.19,  -1,  1.19. 

10.  -  1.88,  2  imag. 

11.  -3.61,  -.87,  .48. 

12.  -  2,  .6,  2. 

13.  -  1.9,  .76,  3.14. 


14.  -  1,  .5,  3. 

15.  -  1.5,  .5,  4. 

16.  -  2,  .5,  4.5. 

17.  -9.28,  2  imag. 

18.  -  1,  2  ±  i. 

19.  -  4,  -  1  ±  i. 

20.  -  1,  2  ±1.41*. 

21.  -  4,  -  1  ±  2  i. 

22.  4,  1  ±  1.41  i. 

23.  -3,  1  ±1.78*. 

24.  -3.84,  1.42  ±1.37  i. 

25.  l,4±2i. 


1.    (a)  5.26. 


Exercise  18.    Page  58 
(6)  1.85.  (c)  1. 


Exercise  19.    Page  61 


-2,  1.66,  2  imag. 
-4,  -1,2,3. 
—  2,  1,  2  imag. 
-4,  -1.83,  2,  3.83. 
-2,  -1,1,2. 


1. 
2. 
3. 
4. 
5. 
6.    1,  3,  2  imag. 


7.  -  2.69,  -  .63,  2  imag. 

8.  1,  2.76,  2  imag. 

9.  -  2.73,  -  1,  .73,  3. 

10.  -2.22,  2.54,  2  imag. 

11.  -2.62,  -.38,  1,  2. 

12.  -3,-2,  1,4. 


ANSWERS 

Exercise  20. 

Pages  62, 63 

1. 

2. 
3. 

4. 
5. 

-6,-2,  1,  7. 
-5,-3,  1,  7. 
-4,  -2,1,5. 
-  5,  -  3,  2,  6. 
-8,-2,1,9. 

6.  -8,-2,  3,  7. 

7.  -6,-4,  3,  7. 

8.  -7,-3,  3,  7. 

9.  -  7,  -  2,  3,  6. 
10.    -6,  -3,  2,7. 

Exercise  21.    Page  65 

1. 
2. 
3. 

4. 
5. 

-  3,  -  1,  2  ±  i. 
-3,  -1,2  ±1.41  i. 
-1,3,  -1±  1.411 
- 1,  3,  - 1  ±  i. 
1.47,  2,  -  1.73  ±  1.04  i. 

6.  -1.82,  1,  .41  ±2.31  i. 

7.  1,  3.61,  -  2.31  ±  1.15  i 

8.  -  4,  2,  1  ±  i. 

9.  -  3,  1,  1  ±  i. 
10.    -  3,  —  1,  2  ±  i. 

93 


Exercise  22.    Page  68 

1.  15.  3.   227.  5.   45.07. 

2.  69.  4.    89.94. 


Exercise  23.    Page  71 

1. 

-5,-2,  1,2.                                 7.    -5,  -3,  1,  4. 

2. 

-3,  -1,  2,6.                                 8.    -7,  -4,  1,  2. 

3. 

-3,  1,4,6.                                     9.    1,2,  3,  4. 

4. 

-5,-3,-1,1.                           10.    -4,-2,1,2. 

5. 

-4,-2,  0,  2.                               11.    -  2.41,  .41,  2,  4. 

6. 

-3,  -1,1,5. 

Exercise  24.    Pages  75,  76,  77 

1. 

6  hrs. ;  18  miles.     5.   2J  hrs.  ;  7£  miles.                   9.   2  :  10.9. 

2. 

18  miles.                  6.   2  :  50.                                      10.   3 :  49.1. 

3. 

4J  hrs.                     7.    (a)  3.  (6)  3.  (c)  4.  (d)  4.     11.   6  :  16.4,  6  :  49.1 

4. 

3f  hrs.                     8.   4.                                             12.   2  hrs. 

13.   1.28  hrs.                            14.    (a)  72.2  meters. 

(6)  155  meters. 

ELEMENTARY  ALGEBRA 

By  ARTHUR  SCHULTZE,  Assistant  Professor  of  Mathematics,  New  York  Univer- 
sity, Head  of  the  Mathematical  Department,  High  School  of  Commerce, 
New  York  City.     i2mo.     Half  leather,    xi  +  373  pages.     $1.10  net. 

The  treatment  of  elementary  algebra  here  is  simple  and  practical,  without 
the  sacrifice  of  scientific  accuracy  and  thoroughness.  Particular  care  has  been 
bestowed  upon  those  chapters  which  in  the  customary  courses  offer  the  great- 
est difficulties  to  the  beginner,  especially  Problems  and  Factoring.  The  intro- 
duction into  Problem  Work  is  very  much  simpler  and  more  natural  than  the 
methods  given  heretofore.  In  Factoring,  comparatively  few  methods  are 
given,  but  these  few  are  treated  so  thoroughly  and  are  illustrated  by  so  many 
varied  examples  that  the  student  will  be  much  better  prepared  for  further 
work,  than  by  the  superficial  study  of  a  great  many  cases.  The  Exercises  are 
very  numerous  and  well  graded;  there  is  a  sufficient  number  of  easy  examples 
of  each  kind  to  enable  the  weakest  students  to  do  some  work.  A  great  many 
examples  are  taken  from  geometry,  physics,  and  commercial  life,  but  none  of 
the  introduced  illustrations  is  so  complex  as  to  require  the  expenditure  of 
time  for  the  teaching  of  physics  or  geometry.  To  meet  the  requirements  of 
the  College  Entrance  Examination  Board,  proportions  and  graphical  methods 
are  introduced  into  the  first  year's  course,  but  the  work  in  the  latter  subject 
has  been  so  arranged  that  teachers  who  wish  a  shorter  course  may  omit  it. 


ADVANCED  ALGEBRA 

By    ARTHUR    SCHULTZE,    Ph.D.        i2mo.        Half  leather.        xiv  +  562    pages. 
$1.25  net. 

The  Advanced  Algebra  is  an  amplification  of  the  Elementary.  All  subjects 
not  now  required  for  admission  by  the  College  Entrance  Examination  Board 
have  been  omitted  from  the  present  volume,  save  Inequalities,  which  has  been 
retained  to  serve  as  a  basis  for  higher  work.  The  more  important  subjects 
which  have  been  omitted  from  the  body  of  the  work  —  Indeterminate  Equa- 
tions, Logarithms,  etc.  —  have  been  relegated  to  the  Appendix,  so  that  the 
book  is  a  thoroughly  practical  and  comprehensive  text-book.  The  author 
has  emphasized  Graphical  Methods  more  than  is  usual  in  text-books  of  this 
grade,  and  the  Summation  of  Species  is  here  presented  in  a   novel    form. 


THE    MACMILLAN    COMPANY 

PUBLISHERS,  64-66  FIFTH  AVENUE,  NEW  YORK 


PLANE  AND   SOLID   GEOMETRY 

By  Arthur   Schultze  and   F.   L.  Sevenoak.     i2mo.     Half  leather.    xii  + 
370  pages.    $1.10  net. 


PLANE  GEOMETRY 

Separate.    i2mo.    Cloth,    xii  +  233  pages.    80  cents  net. 

This  Geometry  introduces  the  student  systematically  to  the  solution  of  geo- 
metrical exercises.  It  provides  a  course  which  stimulates  him  to  do  original 
work  and,  at  the  same  time,  guides  him  in  putting  forth  his  efforts  to  the  best 
advantage. 

The  Schultze  and  Sevenoak  Geometry  is  in  use  in  a  large  number  of  the 
leading  schools  of  the  country.  Attention  is  invited  to  the  following  impor- 
tant features :  I.  Preliminary  Propositions  are  presented  in  a  simple  manner  ; 
2.  The  numerous  and  well-graded  Exercises — more  than  1200  in  number  in 
the  complete  book.  These  are  introduced  from  the  beginning  ;  3.  State- 
ments from  which  General  Principles  may  be  obtained  are  inserted  in  the 
Exercises,  under  the  heading  "  Remarks";  4.  Proofs  that  are  special  cases 
of  general  principles  obtained  from  the  Exercises  are  not  given  in  detail. 
Hints  as  to  the  manner  of  completing  the  work  are  inserted  ;  5.  The  Order 
of  Propositions  has  a  distinct  pedagogical  value.  Propositions  easily  under- 
stood are  given  first  and  more  difficult  ones  follow ;  6.  The  Analysis  of 
Problems  and  of  Theorems  is  more  concrete  and  practical  than  in  any  other 
text-book  in  Geometry ;  7.  Many  proofs  are  presented  in  a  simpler  and 
more  direct  manner  than  in  most  text-books  in  Geometry  ;  8.  Difficult  Prop- 
ositions are  made  somewhat  easier  by  applying  simple  ATotation ;  9.  The 
Algebraic  Solution  of  Geometrical  Exercises  is  treated  in  the  Appendix  to  the 
Plane  Geometry ;  10.  Pains  have  been  taken  to  give  Excellent  Figures 
throughout  the  book. 


KEY  TO   THE  EXERCISES 

In     Schultze     and    Sevenoak's     Plane     and    Solid    Geometry.      By    Arthur 
SCHULTZE,  Ph.D.     i2mo.     Cloth.     200  pages.     #1.10  net. 

This  key  will  be  helpful  to  teachers  who  cannot  give  sufficient  time  to  the 
solution  of  the  exercises  in  the  text-book.  Most  solutions  are  merely  out- 
lines, and  no  attempt  has  been  made  to  present  these  solutions  in  such  form 
that  they  can  be  used  as  models  for  class-room  work. 


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